Phương pháp đổi biến số tìm nguyên hàm

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A. Phương pháp chung
Biết $\int f (u)du = F(u) + C$. Tính $\int f (u(x)).u'(x)dx.$
+ Đặt $t = u(x) \Rightarrow dt = u'(x)dx.$
+ Khi đó: $\int f (u(x)).u'(x)dx$ $ = \int f (t)dt = F(t) + C$ $ = F(u(x)) + C.$

B. Các dạng toán và cách giải
Dạng 1. Nguyên hàm các hàm số đơn giản
Phương pháp:

Dùng phương pháp đổi biến số và áp dụng kết quả sau: Nếu $\int f (x)dx = F(x) + C$ thì $\int f (ax + b)dx = \frac{1}{a}F(ax + b) + C.$
Bảng nguyên hàm các hàm đơn giản:

$u$ là hàm số theo $x$ Trường hợp đặc biệt: $u = ax + b$
$\int d u = u + C$
$\int {{u^\alpha }} du = \frac{{{u^{\alpha + 1}}}}{{\alpha + 1}} + C$ $(\alpha \ne – 1)$ $\int {{{(ax + b)}^\alpha }} dx$ $ = \frac{1}{a} \cdot \frac{{{{(ax + b)}^{\alpha + 1}}}}{{\alpha + 1}} + C$
$\int {\frac{{du}}{u}} = \ln |u| + C$ $(u(x) \ne 0)$ $\int {\frac{{dx}}{{ax + b}}} $ $ = \frac{1}{a}\ln |ax + b| + C$
$\int {\frac{{du}}{{\sqrt u }}} = 2\sqrt u + C$ $(u(x) > 0)$ $\int {\frac{{du}}{{\sqrt {ax + b} }}} $ $ = \frac{1}{a}2\sqrt {ax + b} + C$

Lưu ý cách phân tích: $\frac{1}{{A(A + k)}} = \frac{1}{k}\left( {\frac{1}{A} – \frac{1}{{A + k}}} \right)$, $\forall k \ne 0.$

Ví dụ 1. Tìm họ các nguyên hàm sau:
a) $\int {\sqrt {{{(5 – 3x)}^5}} } dx$ trên khoảng $\left( { – \infty ;\frac{5}{3}} \right).$
b) $\int {\left( {\frac{1}{{{{(3x + 1)}^3}}} + \frac{2}{{\sqrt {4x – 3} }}} \right)} dx.$
c) $\int {\frac{{dx}}{{9 – 4{x^2}}}} .$

a) Trên khoảng $\left( { – \infty ;\frac{5}{3}} \right)$, ta có:
$\int {\sqrt {{{(5 – 3x)}^5}} } dx = \int {{{(5 – 3x)}^{\frac{5}{2}}}} dx$ $ = – \frac{1}{3}{(5 – 3x)^{\frac{7}{2}}} \cdot \frac{2}{7} + C$ $ = – \frac{2}{{21}}\sqrt {{{(5 – 3x)}^7}} + C.$
b) $\int {\left( {\frac{1}{{{{(3x + 1)}^3}}} + \frac{2}{{\sqrt {4x – 3} }}} \right)} dx$ $ = \int {\frac{1}{{{{(3x + 1)}^3}}}} dx + \int {\frac{2}{{\sqrt {4x – 3} }}} dx$ $ = \int {{{(3x + 1)}^{ – 3}}} dx + \int {\frac{2}{{\sqrt {4x – 3} }}} dx$ $ = – \frac{1}{6} \cdot \frac{1}{{{{(3x + 1)}^2}}} + \sqrt {4x – 3} + C.$
c) $\int {\frac{{dx}}{{9 – 4{x^2}}}} $ $ = – \int {\frac{{dx}}{{(2x – 3)(2x + 3)}}} $ $ = – \frac{1}{6}\int {\left( {\frac{1}{{2x – 3}} – \frac{1}{{2x + 3}}} \right)} dx$ $ = – \frac{1}{6} \cdot \left( {\frac{1}{2}\ln |2x – 3| – \frac{1}{2}\ln |2x + 3|} \right) + C$ $ = \frac{1}{{12}}\ln \left| {\frac{{2x + 3}}{{2x – 3}}} \right| + C.$

Ví dụ 2. Tìm họ các nguyên hàm sau:
a) $\int {\frac{{{x^3}}}{{{x^8} – 8}}} dx.$
b) $\int {\frac{{{x^2}}}{{{{(1 – x)}^{100}}}}} dx.$
c) $\int {\frac{{{x^2} – 1}}{{{x^4} + 1}}} dx.$

a) Đặt $t = {x^4} \Rightarrow dt = 4{x^3}dx$ $ \Rightarrow {x^3}dx = \frac{{dt}}{4}.$
$ \Rightarrow \int {\frac{{{x^3}}}{{{x^8} – 8}}} dx = \frac{1}{4}\int {\frac{{dt}}{{{t^2} – 8}}} $ $ = \frac{1}{4}\int {\frac{1}{{(t – 2\sqrt 2 )(t + 2\sqrt 2 )}}} dt$ $ = \frac{1}{{16\sqrt 2 }}\int {\left( {\frac{1}{{t – 2\sqrt 2 }} – \frac{1}{{t + 2\sqrt 2 }}} \right)} dt$ $ = \frac{1}{{16\sqrt 2 }}\ln \left| {\frac{{t – 2\sqrt 2 }}{{t + 2\sqrt 2 }}} \right| + C$ $ = \frac{1}{{16\sqrt 2 }}\ln \left| {\frac{{{x^4} – 2\sqrt 2 }}{{{x^4} + 2\sqrt 2 }}} \right| + C.$
b) Đặt $t = 1 – x$ $ \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{dt = – dx \Rightarrow dx = – dt}\\
{x = 1 – t}
\end{array}} \right.$
$ \Rightarrow \int {\frac{{{x^2}}}{{{{(1 – x)}^{100}}}}} dx$ $ = \int {\frac{{{{(1 – t)}^2}dt}}{{{t^{100}}}}} $ $ = \int {\frac{{1 – 2t + {t^2}}}{{{t^{100}}}}} dt$ $ = \int {\left( {\frac{1}{{{t^{100}}}} – \frac{2}{{{t^{99}}}} + \frac{1}{{{t^{98}}}}} \right)} dt$ $ = – \frac{1}{{99}} \cdot \frac{1}{{{t^{99}}}} + \frac{2}{{98}} \cdot \frac{1}{{{t^{98}}}} – \frac{1}{{97}} \cdot \frac{1}{{{t^{97}}}} + C$ $ = – \frac{1}{{99}} \cdot \frac{1}{{{{(1 – x)}^{99}}}} + \frac{2}{{98}},\frac{1}{{{{(1 – x)}^{98}}}} – \frac{1}{{97}} \cdot \frac{1}{{{{(1 – x)}^{97}}}} + C.$
c) $\int {\frac{{{x^2} – 1}}{{{x^4} + 1}}} dx = \int {\frac{{1 – \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}}}}} dx.$
Đặt $t = x + \frac{1}{x}$ $ \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{dt = \left( {1 – \frac{1}{{{x^2}}}} \right)dx}\\
{{t^2} = {x^2} + \frac{1}{{{x^2}}} + 2 \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {t^2} – 2}
\end{array}} \right.$
$ \Rightarrow \int {\frac{{{x^2} – 1}}{{{x^4} + 1}}} dx = \int {\frac{{dt}}{{{t^2} – 2}}} $ $ = \int {\frac{{dt}}{{(t – \sqrt 2 )(t + \sqrt 2 )}}} $ $ = \frac{1}{{2\sqrt 2 }}\int {\left( {\frac{1}{{t – \sqrt 2 }} – \frac{1}{{t + \sqrt 2 }}} \right)} dt$ $ = \frac{1}{{2\sqrt 2 }}\ln \left| {\frac{{t – \sqrt 2 }}{{t + \sqrt 2 }}} \right| + C$ $ = \frac{1}{{2\sqrt 2 }}\ln \left| {\frac{{x + \frac{1}{x} – \sqrt 2 }}{{x + \frac{1}{x} + \sqrt 2 }}} \right| + C$ $ = \frac{1}{{2\sqrt 2 }}\ln \left| {\frac{{{x^2} – x\sqrt 2 + 1}}{{{x^2} + x\sqrt 2 + 1}}} \right| + C.$

Ví dụ 3. Tìm các họ nguyên hàm sau:
a) $\int {\frac{{6x}}{{\sqrt[3]{{{x^2} + 4}}}}} dx.$
b) $\int {\frac{{dx}}{{x.\ln x.\ln (\ln x)}}} .$
c) $\int {\frac{{\sqrt {1 + \ln x} }}{x}} dx.$

a) Đặt: $t = \sqrt[3]{{{x^2} + 4}}$ $ \Rightarrow {t^3} = {x^2} + 4$ $ \Rightarrow 3{t^2}dt = 2xdx$ $ \Rightarrow 6xdx = 9{t^2}dt.$
$ \Rightarrow \int {\frac{{6x}}{{\sqrt[3]{{{x^2} + 4}}}}} dx$ $ = \int {\frac{{9{t^2}dt}}{t}} = \int 9 tdt$ $ = \frac{9}{2}{t^2} + C$ $ = \frac{9}{2}{\left( {\sqrt[3]{{{x^2} + 4}}} \right)^2} + C.$
b) Đặt $t = \ln (\ln x)$ $ \Rightarrow dt = \frac{1}{{x\ln x}}dx.$
$ \Rightarrow \int {\frac{{dx}}{{x.\ln x.\ln (\ln x)}}} $ $ = \int {\frac{{dt}}{t}} = \ln |t| + C$ $ = \ln |\ln (\ln x)| + C.$
c) Đặt $t = \sqrt {1 + \ln x} $ $ \Rightarrow {t^2} = 1 + \ln x$ $ \Rightarrow 2 tdt = \frac{1}{x}dx.$
$ \Rightarrow \int {\frac{{\sqrt {1 + \ln x} }}{x}} dx = \int 2 {t^2}dt$ $ = \frac{2}{3}{t^3} + C$ $ = \frac{2}{3}\sqrt {{{(1 + \ln x)}^3}} + C.$

Dạng 2. Nguyên hàm của hàm số mũ
Phương pháp
:

$u$ là hàm số theo $x$ Trường hợp đặc biệt: $u = ax + b$ $(a \ne 0)$
$\int {{e^u}} du = {e^u} + C$ $\int {{e^{ax + b}}} dx = \frac{1}{a}.{e^{ax + b}} + C$
$\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C$ $(0 < a \ne 1)$ $\int {{a^{mx + n}}} du = \frac{{{a^{mx + n}}}}{{m.\ln a}} + C$ $(0 < a \ne 1, m \ne 0)$

Ví dụ 1. Tìm các họ nguyên hàm sau:
a) $\int {\left( {{e^{2x + 4}} – {e^{2 – 5x}}} \right)} dx.$
b) $\int {(x + 1)} {e^{{x^2} + 2x}}dx.$
c) $\int {\frac{{dx}}{{{e^x} – {e^{ – x}}}}} .$
d) $\int {\frac{{{e^{2x}}dx}}{{1 + {e^x}}}} .$

a) $\int {\left( {{e^{2x + 4}} – {e^{2 – 5x}}} \right)} dx$ $ = \frac{1}{2}{e^{2x + 4}} + \frac{1}{5}{e^{2 – 5x}} + C.$
b) Đặt: $t = {x^2} + 2x$ $ \Rightarrow dt = (2x + 2)dt$ $ \Rightarrow (x + 1)dx = \frac{{dt}}{2}.$
$ \Rightarrow \int {(x + 1)} {e^{{x^2} + 2x}}dx$ $ = \int {{e^t}} \frac{{dt}}{2} = \frac{{{e^t}}}{2} + C$ $ = \frac{{{e^{{x^2} + 2x}}}}{2} + C.$
c) $\int {\frac{{dx}}{{{e^x} – {e^{ – x}}}}} = \int {\frac{{{e^x}dx}}{{{e^{2x}} – 1}}} .$
Đặt: $t = {e^x} \Rightarrow dt = {e^x}dx.$
$ \Rightarrow \int {\frac{{{e^x}dx}}{{{e^{2x}} – 1}}} = \int {\frac{{dt}}{{{t^2} – 1}}} $ $ = \int {\frac{{dt}}{{(t – 1)(t + 1)}}} $ $ = \frac{1}{2}\int {\left( {\frac{1}{{t – 1}} – \frac{1}{{t + 1}}} \right)} dt$ $ = \frac{1}{2}\ln \left| {\frac{{t – 1}}{{t + 1}}} \right| + C$ $ = \frac{1}{2}\ln \left| {\frac{{{e^x} – 1}}{{{e^x} + 1}}} \right| + C.$
d) Đặt: $t = 1 + {e^x} \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{dt = {e^x}dx}\\
{{e^x} = t – 1}
\end{array}} \right.$
$ \Rightarrow \int {\frac{{{e^{2x}}dx}}{{1 + {e^x}}}} = \int {\frac{{{e^x} \cdot {e^x}dx}}{{1 + {e^X}}}} $ $ = \int {\frac{{(t – 1)dt}}{t}} = \int {\left( {1 – \frac{1}{t}} \right)} dt$ $ = t – \ln |t| + C$ $ = 1 + {e^x} – \ln \left( {1 + {e^x}} \right) + C.$

Ví dụ 2. Tìm các họ nguyên hàm sau:
a) $\int {\frac{{{6^x}}}{{{9^x} – {4^x}}}} dx.$
b) $\int {\frac{{{e^{2x}}dx}}{{\sqrt {1 + {e^x}} + \sqrt {1 – {e^x}} }}} .$
c) $\int {{e^{\sin x + 1}}} \cos xdx.$

a) $\int {\frac{{{6^x}}}{{{9^x} – {4^x}}}} dx$ $ = \int {\frac{{{{\left( {\frac{3}{2}} \right)}^x}}}{{{{\left( {\frac{3}{2}} \right)}^{2x}} – 1}}} dx.$
Đặt: $t = {\left( {\frac{3}{2}} \right)^x}$ $ \Rightarrow dt = {\left( {\frac{3}{2}} \right)^x}\ln \frac{3}{2}dx$ $ \Rightarrow {\left( {\frac{3}{2}} \right)^x}dx = \frac{{dt}}{{\ln \frac{3}{2}}}.$
$ \Rightarrow \int {\frac{{{{\left( {\frac{3}{2}} \right)}^x}}}{{{{\left( {\frac{3}{2}} \right)}^{2x}} – 1}}} dx$ $ = \frac{1}{{\ln \frac{3}{2}}}\int {\frac{{dt}}{{{t^2} – 1}}} $ $ = \frac{1}{{\ln \frac{3}{2}}}\int {\frac{{dt}}{{(t – 1)(t + 1)}}} $ $ = \frac{1}{{2\ln \frac{3}{2}}}\int {\left( {\frac{1}{{t – 1}} – \frac{1}{{t + 1}}} \right)} dt$ $ = \frac{1}{{2\ln \frac{3}{2}}}\ln \left| {\frac{{t – 1}}{{t + 1}}} \right| + C$ $ = \frac{1}{{2\ln \frac{3}{2}}}\ln \left| {\frac{{{{\left( {\frac{3}{2}} \right)}^x} – 1}}{{{{\left( {\frac{3}{2}} \right)}^x} + 1}}} \right| + C$ $ = \frac{1}{{2\ln \frac{3}{2}}}\ln \left| {\frac{{{3^x} – {2^x}}}{{{3^x} + {2^x}}}} \right| + C.$
b) Đặt: $t = {e^x} \Rightarrow dt = {e^x}dx.$
$ \Rightarrow \int {\frac{{{e^{2x}}dx}}{{\sqrt {1 + {e^x}} + \sqrt {1 – {e^x}} }}} $ $ = \int {\frac{{ tdt }}{{\sqrt {1 + t} + \sqrt {1 – t} }}} $ $ = \int {\frac{{\sqrt {1 + t} – \sqrt {1 – t} }}{2}} dt$ $ = \frac{1}{3}\left( {\sqrt {{{(1 + t)}^3}} + \sqrt {{{(1 – t)}^3}} } \right) + C$ $ = \frac{1}{3}\left( {\sqrt {{{\left( {1 + {e^x}} \right)}^3}} + \sqrt {{{\left( {1 – {e^x}} \right)}^3}} } \right) + C.$
c) Đặt $t = \sin x + 1$ $ \Rightarrow dt = \cos xdx.$
$ \Rightarrow \int {{e^{\sin x + 1}}} \cos xdx$ $ = \int {{e^t}} dt = {e^t} + C = {e^{\sin x + 1}} + C.$

Dạng 3. Nguyên hàm của hàm phân thức hữu tỉ
Phương pháp: $y = f(x) = \frac{{P(x)}}{{Q(x)}}$ ($P(x),Q(x)$ là hàm đa thức).
+ Nếu bậc $P(x)$ < bậc $Q(x)$: Ta biến đổi $f(x)$ về tổng (hiệu) các hàm số đơn giản rồi tìm nguyên hàm.
+ Nếu bậc $P(x)$ $≥$ bậc $Q(x)$: Ta thực hiện phép chia đa thức $P(x)$ cho $Q(x)$ rồi biến đổi $f(x)$ về tổng (hiệu) các hàm số đơn giản rồi tìm nguyên hàm.
Lưu ý cách phân tích sau:
+ Nếu tam thức $a{x^2} + bx + c$ có hai nghiệm ${x_1},{x_2}$ thì: $a{x^2} + bx + c$ $ = a\left( {x – {x_1}} \right)\left( {x – {x_2}} \right).$
+ $\int {\frac{1}{{a\left( {x – {x_1}} \right)\left( {x – {x_2}} \right)}}} dx$ $ = \frac{1}{{a\left( {{x_2} – {x_1}} \right)}}\int {\left[ {\frac{1}{{x – {x_2}}} – \frac{1}{{x – {x_1}}}} \right]} dx.$
+ Cho $P(x) = {a_n}{x^n} + {a_{n – 1}}{x^{n – 1}} + … + {a_1}x + {a_0}$ và $Q(x) = {b_n}{x^n} + {b_{n – 1}}{x^{n – 1}} + \ldots + {b_1}x + {b_0}.$ Khi đó: $P(x) = Q(x)$ $(\forall x \in R)$ $ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{{a_n} = {b_n}}\\
{{a_{n – 1}} = {b_{n – 1}}}\\
{ \ldots \ldots }\\
{{a_1} = {b_1}}\\
{{a_0} = {b_0}}
\end{array}} \right.$ (Phương pháp đồng nhất thức).

Ví dụ 1. Tìm các nguyên hàm sau:
a) $\int {\frac{{dx}}{{\left( {{x^2} – 4x + 3} \right)\left( {{x^2} – 4x + 4} \right)}}} .$
b) $\int {\frac{{dx}}{{x(x + 1)\left( {{x^2} + x – 2} \right)}}} .$

a) $\int {\frac{{dx}}{{\left( {{x^2} – 4x + 3} \right)\left( {{x^2} – 4x + 4} \right)}}} $ $ = \int {\frac{{dx}}{{{x^2} – 4x + 3}}} – \int {\frac{{dx}}{{{x^2} – 4x + 4}}} $ $ = \int {\frac{{dx}}{{(x – 1)(x – 3)}}} – \int {\frac{{dx}}{{{{(x – 2)}^2}}}} $ $ = \frac{1}{2}\int {\left( {\frac{1}{{x – 3}} – \frac{1}{{x – 1}}} \right)} dx – \int {\frac{{dx}}{{{{(x – 2)}^2}}}} $ $ = \frac{1}{2}\ln \left| {\frac{{x – 3}}{{x – 1}}} \right| + \frac{1}{{x – 2}} + C.$
b) $\int {\frac{{dx}}{{x(x + 1)\left( {{x^2} + x – 2} \right)}}} $ $ = \int {\frac{{dx}}{{\left( {{x^2} + x} \right)\left( {{x^2} + x – 2} \right)}}} $ $ = \frac{1}{2}\int {\left( {\frac{1}{{{x^2} + x – 2}} – \frac{1}{{{x^2} + x}}} \right)} dx$ $ = \frac{1}{2}\int {\frac{{dx}}{{(x – 1)(x + 2)}}} – \frac{1}{2}\int {\frac{{dx}}{{x(x + 1)}}} $ $ = \frac{1}{6}\int {\left( {\frac{1}{{x – 1}} – \frac{1}{{x + 2}}} \right)} dx – \frac{1}{2}\int {\left( {\frac{1}{x} – \frac{1}{{x + 1}}} \right)} dx$ $ = \frac{1}{6}\ln \left| {\frac{{x – 1}}{{x + 2}}} \right| – \frac{1}{2}\ln \left| {\frac{x}{{x + 1}}} \right| + C.$

Ví dụ 2. Tìm các nguyên hàm sau:
a) $\int {\frac{4}{{{x^4} – 7{x^2} + 6}}} dx.$
b) $\int {\frac{{2x – 3}}{{3{x^2} – 2x – 1}}} dx.$
c) $\int {\frac{{{x^5}}}{{{x^6} – {x^3} – 2}}} dx.$
d) $\int {\frac{{{x^4}}}{{{x^4} – 5{x^2} + 4}}} dx.$

a) $\int {\frac{4}{{{x^4} – 7{x^2} + 6}}} dx$ $ = \int {\frac{4}{{\left( {{x^2} – 1} \right)\left( {{x^2} – 6} \right)}}} dx$ $ = \frac{4}{5}\int {\left( {\frac{1}{{{x^2} – 6}} – \frac{1}{{{x^2} – 1}}} \right)} dx$ $ = \frac{4}{5}\int {\frac{1}{{{x^2} – 6}}} dx – \frac{4}{5}\int {\frac{1}{{{x^2} – 1}}} dx$ $ = \frac{4}{5}\int {\frac{1}{{(x – \sqrt 6 )(x + \sqrt 6 )}}} dx$ $ – \frac{4}{5}\int {\frac{1}{{(x – 1)(x + 1)}}} dx$ $ = \frac{2}{{5\sqrt 6 }}\int {\left( {\frac{1}{{x – \sqrt 6 }} – \frac{1}{{x + \sqrt 6 }}} \right)} dx$ $ – \frac{2}{5}\int {\left( {\frac{1}{{x – 1}} – \frac{1}{{x + 1}}} \right)} dx$ $ = \frac{2}{{5\sqrt 6 }}\ln \left| {\frac{{x – \sqrt 6 }}{{x + \sqrt 6 }}} \right|$ $ – \frac{2}{5}\ln \left| {\frac{{x – 1}}{{x + 1}}} \right| + C.$
b) $\int {\frac{{2x – 3}}{{3{x^2} – 2x – 1}}} dx$ $ = \frac{1}{3}\int {\frac{{6x – 2 – 7}}{{3{x^2} – 2x – 1}}} dx$ $ = \frac{1}{3}\int {\frac{{6x – 2}}{{3{x^2} – 2x – 1}}} dx$ $ – \frac{7}{3}\int {\frac{{dx}}{{3{x^2} – 2x – 1}}} $ $ = \frac{1}{3}\int {\frac{{{{\left( {3{x^2} – 2x – 1} \right)}^\prime }}}{{3{x^2} – 2x – 1}}} dx$ $ – \frac{7}{3}\int {\frac{{dx}}{{3(x – 1)\left( {x + \frac{1}{3}} \right)}}} $ $ = \frac{1}{3}\int {\frac{{{{\left( {3{x^2} – 2x – 1} \right)}^\prime }}}{{3{x^2} – 2x – 1}}} dx$ $ – \frac{7}{9} \cdot \frac{3}{4}\int {\left( {\frac{1}{{x – 1}} – \frac{1}{{x + \frac{1}{3}}}} \right)} dx$ $ = \frac{1}{3}\ln \left| {3{x^2} – 2x – 1} \right|$ $ – \frac{7}{{12}}\ln \left| {\frac{{x – 1}}{{x + \frac{1}{3}}}} \right| + C$ $ = \frac{1}{3}\ln \left| {3{x^2} – 2x – 1} \right|$ $ – \frac{7}{{12}}\ln \left| {\frac{{3(x – 1)}}{{3x + 1}}} \right| + C.$
c)
Cách 1: $\int {\frac{{{x^5}}}{{{x^6} – {x^3} – 2}}} dx$ $ = \int {\frac{{{x^5}}}{{\left( {{x^3} + 1} \right)\left( {{x^3} – 2} \right)}}} dx$ $ = \frac{1}{3}\int {\left( {\frac{{{x^5}}}{{{x^3} – 2}} – \frac{{{x^5}}}{{{x^3} + 1}}} \right)} dx$ $ = \frac{1}{3}\int {\frac{{{x^5}}}{{{x^3} – 2}}} dx – \frac{1}{3}\int {\frac{{{x^5}}}{{{x^3} + 1}}} dx$ $ = A – B.$
Tính $A$:
Đặt: $t = {x^3} – 2$ $ \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{dt = 3{x^2}dx \Rightarrow {x^2}dx = \frac{{dt}}{3}}\\
{{x^3} = t + 2}
\end{array}} \right.$
$ \Rightarrow A = \frac{1}{3}\int {\frac{{{x^5}}}{{{x^3} – 2}}} dx$ $ = \frac{1}{3}\int {\frac{{(t + 2)dt}}{{3t}}} $ $ = \frac{1}{9}\int {\left( {1 + \frac{2}{t}} \right)} dt$ $ = \frac{1}{9}(t + 2\ln |t|) + C$ $ = \frac{1}{9}\left( {{x^3} – 2 + 2\ln \left| {{x^3} – 2} \right|} \right) + C.$
Tính $B$:
Đặt: $t = {x^3} + 1$ $ \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{dt = 3{x^2}dx \Rightarrow {x^2}dx = \frac{{dt}}{3}}\\
{{x^3} = t – 1}
\end{array}} \right.$
$ \Rightarrow B = \frac{1}{3}\int {\frac{{{x^5}}}{{{x^3} + 1}}} dx$ $ = \frac{1}{3}\int {\frac{{(t – 1)dt}}{{3t}}} $ $ = \frac{1}{9}\int {\left( {1 – \frac{1}{t}} \right)} dt$ $ = \frac{1}{9}(t – \ln |t|) + C$ $ = \frac{1}{9}\left( {{x^3} + 1 – \ln \left| {{x^3} + 1} \right|} \right) + C.$
Suy ra: $\int {\frac{{{x^5}}}{{{x^6} – {x^3} – 2}}} dx$ $ = \frac{1}{9}\left( {{x^3} – 2 + 2\ln \left| {{x^3} – 2} \right|} \right)$ $ – \frac{1}{9}\left( {{x^3} + 1 – \ln \left| {{x^3} + 1} \right|} \right) + C$ $ = \frac{1}{9}\ln \left( {\left| {{x^3} + 1} \right|{{\left( {{x^3} – 2} \right)}^2}} \right) + C.$
Cách 2:
$\int {\frac{{{x^5}}}{{{x^6} – {x^3} – 2}}} dx$ $ = \int {\frac{{{x^3}.{x^2}}}{{{{\left( {{x^3} – \frac{1}{2}} \right)}^2} – \frac{9}{4}}}} dx.$
Đặt: $t = {x^3} – \frac{1}{2}$ $ \Rightarrow dt = 3{x^2}dx$ $ \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{{x^2}dx = \frac{{dt}}{3}}\\
{{x^3} = t + \frac{1}{2}}
\end{array}} \right.$
$ \Rightarrow \int {\frac{{{x^3}{x^2}}}{{{{\left( {{x^3} – \frac{1}{2}} \right)}^2} – \frac{9}{4}}}} dx$ $ = \frac{1}{3}\int {\frac{{\left( {t + \frac{1}{2}} \right)dt}}{{{t^2} – \frac{9}{4}}}} $ $ = \frac{1}{3}\int {\frac{t}{{{t^2} – \frac{9}{4}}}} dt$ $ + \frac{2}{3}\int {\frac{1}{{(2t – 3)(2t + 3)}}} dt$ $ = \frac{1}{6}\int {\frac{{2t}}{{{t^2} – \frac{9}{4}}}} dt$ $ + \frac{1}{9}\int {\left( {\frac{1}{{2t – 3}} – \frac{1}{{2t + 3}}} \right)} dt$ $ = \frac{1}{6}\ln \left| {{t^2} – \frac{9}{4}} \right|$ $ + \frac{1}{{18}}\ln \left| {\frac{{2t – 3}}{{2t + 3}}} \right| + C$ $ = \frac{1}{2}\ln \left| {{x^6} – {x^3} – 2} \right|$ $ + \frac{1}{{18}}\ln \left| {\frac{{{x^3} – 2}}{{{x^3} + 1}}} \right| + C.$
d) $\int {\frac{{{x^4}}}{{{x^4} – 5{x^2} + 4}}} dx$ $ = \int {\frac{{{x^4}}}{{\left( {{x^2} – 1} \right)\left( {{x^2} – 4} \right)}}} dx$ $ = \frac{1}{3}\int {\left( {\frac{{{x^4}}}{{{x^2} – 4}} – \frac{{{x^4}}}{{{x^2} – 1}}} \right)} dx$ $ = \frac{1}{3}\int {\left[ {\left( {{x^2} + 4 + \frac{{16}}{{{x^2} – 4}}} \right) – \left( {{x^2} + 1 + \frac{1}{{{x^2} – 1}}} \right)} \right]} dx$ $ = \frac{1}{3}\int {\left( {3 – \frac{1}{{{x^2} – 1}} + \frac{{16}}{{{x^2} – 4}}} \right)} dx$ $ = \frac{1}{3}\int 3 dx – \frac{1}{3}\int {\frac{1}{{(x – 1)(x + 1)}}} dx$ $ + \frac{1}{3}\int {\frac{{16}}{{(x – 2)(x + 2)}}} dx$ $ = \frac{1}{3}\int 3 dx – \frac{1}{6}\int {\left( {\frac{1}{{x – 1}} – \frac{1}{{x + 1}}} \right)} dx$ $ + \frac{4}{3}\int {\left( {\frac{1}{{x – 2}} – \frac{1}{{x + 2}}} \right)} dx$ $ = x – \frac{1}{6}\ln \left| {\frac{{x – 1}}{{x + 1}}} \right| + \frac{4}{3}\ln \left| {\frac{{x – 2}}{{x + 2}}} \right| + C.$

Ví dụ 3.
a) Cho $f(x) = \frac{1}{{x(x – 1)(x – 2)}}.$ Tính $\int f (x)dx.$
b) Cho $f(x) = \frac{1}{{(x + 1){{(x – 1)}^2}}}.$ Tính $\int f (x)dx.$

a)
Cách 1: $f(x) = \frac{1}{{x(x – 1)(x – 2)}}$ $ = \frac{1}{x}\left( {\frac{1}{{x – 2}} – \frac{1}{{x – 1}}} \right)$ $ = \frac{1}{{x(x – 2)}} – \frac{1}{{x(x – 1)}}$ $ = \frac{1}{2}\left( {\frac{1}{{x – 2}} – \frac{1}{x}} \right) – \left( {\frac{1}{{x – 1}} – \frac{1}{x}} \right)$ $ = \frac{1}{2} \cdot \frac{1}{{x – 2}} – \frac{1}{{x – 1}} + \frac{1}{2} \cdot \frac{1}{x}.$
$ \Rightarrow \int f (x)dx$ $ = \frac{1}{2}\ln |x – 2| – \ln |x – 1| + \frac{1}{2}\ln |x| + C$ $ = \frac{1}{2}\ln |x(x – 2)| – \ln |x – 1| + C.$
Cách 2: Dùng phương pháp đồng nhất thức:
$f(x) = \frac{1}{{x(x – 1)(x – 2)}}$ $ = \frac{A}{x} + \frac{B}{{x – 1}} + \frac{C}{{x – 2}}$ $ = \frac{{A(x – 1)(x – 2) + Bx(x – 2) + Cx(x – 1)}}{{x(x – 1)(x – 2)}}$ $ = \frac{{(A + B + C){x^2} – (3A + 2B + C)x + 2A}}{{x(x – 1)(x – 2)}}.$
Như vậy, với mọi $x \in R\backslash \{ 0,1,2\} $, ta phải có hệ sau:
$\left\{ {\begin{array}{*{20}{l}}
{A + B + C = 0}\\
{3A + 2B + C = 0}\\
{2A = 1}
\end{array}} \right.$ $ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{A = \frac{1}{2}}\\
{B = – 1}\\
{C = \frac{1}{2}}
\end{array}} \right.$
Vậy: $f(x) = \frac{1}{2} \cdot \frac{1}{x} – \frac{1}{{x – 1}} + \frac{1}{2} \cdot \frac{1}{{x – 2}}.$
$ \Rightarrow \int f (x)dx$ $ = \frac{1}{2}\ln |x| – \ln |x – 1| + \frac{1}{2}\ln |x – 2| + C$ $ = \frac{1}{2}\ln |x(x – 2)| – \ln |x – 1| + C.$
b)
Cách 1: $f(x) = \frac{1}{{(x + 1){{(x – 1)}^2}}}$ $ = \frac{1}{2} \cdot \frac{1}{{x – 1}}\left( {\frac{1}{{x – 1}} – \frac{1}{{x + 1}}} \right)$ $ = \frac{1}{2} \cdot \frac{1}{{{{(x – 1)}^2}}} – \frac{1}{2} \cdot \frac{1}{{(x – 1)(x + 1)}}$ $ = \frac{1}{2} \cdot \frac{1}{{{{(x – 1)}^2}}}$ $ – \frac{1}{4} \cdot \left( {\frac{1}{{x – 1}} – \frac{1}{{x + 1}}} \right).$
$ \Rightarrow \int f (x)dx$ $ = – \frac{1}{2} \cdot \frac{1}{{x – 1}} – \frac{1}{4}\ln \left| {\frac{{x – 1}}{{x + 1}}} \right| + C.$
Cách 2: Phương pháp đồng nhất thức:
$f(x) = \frac{1}{{(x + 1){{(x – 1)}^2}}}$ $ = \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{{(x – 1)}^2}}}$ $ = \frac{{A{{(x – 1)}^2} + (Bx + C)(x + 1)}}{{(x + 1){{(x – 1)}^2}}}$ $ = \frac{{(A + B){x^2} + (B + C – 2A)x + (A + C)}}{{(x + 1){{(x – 1)}^2}}}.$
Như vậy, với mọi $x \in R\backslash \{ – 1,1\} $, ta phải có hệ sau:
$\left\{ {\begin{array}{*{20}{l}}
{A + B = 0}\\
{ – 2A + B + C = 0}\\
{A + C = 1}
\end{array}} \right.$ $ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{A = \frac{1}{4}}\\
{B = – \frac{1}{4}}\\
{C = \frac{3}{4}}
\end{array}} \right.$
$f(x) = \frac{1}{4} \cdot \frac{1}{{x + 1}} – \frac{1}{4} \cdot \frac{{x – 3}}{{{{(x – 1)}^2}}}$ $ = \frac{1}{4} \cdot \frac{1}{{x + 1}} – \frac{1}{4}\frac{{(x – 1) – 2}}{{{{(x – 1)}^2}}}$ $ = \frac{1}{4} \cdot \frac{1}{{x + 1}} – \frac{1}{4} \cdot \frac{1}{{x – 1}} + \frac{1}{2} \cdot \frac{1}{{{{(x – 1)}^2}}}$ $ \Rightarrow \int f (x)dx$ $ = – \frac{1}{4}\ln \left| {\frac{{x – 1}}{{x + 1}}} \right| – \frac{1}{2} \cdot \frac{1}{{x – 1}} + C.$

Dạng 4. Nguyên hàm các hàm căn thức
Phương pháp
: Một số lưu ý khi tính nguyên hàm có chứa dấu căn như sau:
+ Dùng công thức biến đổi căn: $\sqrt[n]{{{u^m}}} = {u^{\frac{m}{n}}}$ $(u > 0).$
+ Khử dấu bằng cách nhân lượng liên hợp.
+ Thường đổi biến: Đặt $t = \sqrt[n]{{u(x)}}$ $ \Rightarrow {t^n} = u(x)$ $ \Rightarrow n.{t^{n – 1}}dt = u'(x)dx.$

$u$ là hàm số theo $x$ Trường hợp đặc biệt: $u=ax+b$
$\int {\frac{{du}}{{\sqrt u }}} = 2\sqrt u + C$ $(u(x) > 0)$ $\int {\frac{{du}}{{\sqrt {ax + b} }}} $ $ = \frac{1}{a}2\sqrt {ax + b} + C$

Ví dụ. Tìm nguyên hàm các hàm số sau:
a) $\int {\frac{{dx}}{{\sqrt {3x + 1} + \sqrt {3x – 1} }}} $ trên khoảng $\left( {\frac{1}{3}; + \infty } \right).$
b) $\int {\frac{{dx}}{{x\sqrt {1 + {x^3}} }}} .$
c) $\int {\frac{x}{{\sqrt {1 + \sqrt x } }}} dx.$
d) $\int {\frac{{xdx}}{{\sqrt {{{\left( {1 + {x^2}} \right)}^3}} }}} .$

a) $\int {\frac{{dx}}{{\sqrt {3x + 1} + \sqrt {3x – 1} }}} $ $ = \frac{1}{2}\int {(\sqrt {3x + 1} – \sqrt {3x – 1} )} dx$ $ = \frac{1}{2}\int {\left[ {{{(3x + 1)}^{\frac{1}{2}}} – {{(3x – 1)}^{\frac{1}{2}}}} \right]} dx$ $ = \frac{1}{9}\left[ {\sqrt {{{(3x + 1)}^3}} – \sqrt {{{(3x – 1)}^3}} } \right] + C.$
b) $\int {\frac{{dx}}{{x\sqrt {1 + {x^3}} }}} $ $ = \int {\frac{{{x^2}dx}}{{{x^3}\sqrt {1 + {x^3}} }}} .$
Đặt: $t = \sqrt {1 + {x^3}} $ $ \Rightarrow {t^2} = 1 + {x^3}$ $ \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{2tdt = 3{x^2}dx \Rightarrow {x^2}dx = \frac{2}{3}{\rm{ }}tdt{\rm{ }}}\\
{{x^3} = {t^2} – 1}
\end{array}} \right.$
$ \Rightarrow \int {\frac{{dx}}{{x\sqrt {1 + {x^3}} }}} $ $ = \frac{2}{3}\int {\frac{{ tdt }}{{\left( {{t^2} – 1} \right).t}}} $ $ = \frac{2}{3}\int {\frac{{dt}}{{(t – 1)(t + 1)}}} $ $ = \frac{1}{3}\int {\left( {\frac{1}{{t – 1}} – \frac{1}{{t + 1}}} \right)} dt$ $ = \frac{1}{3}\ln \left| {\frac{{t – 1}}{{t + 1}}} \right| + C$ $ = \frac{1}{3}\ln \left| {\frac{{\sqrt {1 + {x^3}} – 1}}{{\sqrt {1 + {x^3}} + 1}}} \right| + C.$
c) Đặt: $t = \sqrt {1 + \sqrt x } $ $ \Rightarrow {t^2} = 1 + \sqrt x $ $ \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{2tdt = \frac{1}{{2\sqrt x }}dx \Rightarrow dx = 4t\left( {{t^2} – 1} \right)dt}\\
{\sqrt x = {t^2} – 1}
\end{array}} \right.$
$ \Rightarrow \int {\frac{x}{{\sqrt {1 + \sqrt x } }}} dx$ $ = \int {\frac{{{{\left( {{t^2} – 1} \right)}^2}4t\left( {{t^2} – 1} \right)}}{t}} dt$ $ = 4\int {{{\left( {{t^2} – 1} \right)}^3}} dt$ $ = 4\int {\left( {{t^6} – 3{t^4} + 3{t^2} – 1} \right)} dt$ $ = \frac{4}{7}{t^7} – \frac{{12}}{5}{t^5} + 4{t^3} – 4t + C$ $ = \frac{4}{7}\sqrt {{{(1 + \sqrt x )}^7}} – \frac{{12}}{5}\sqrt {{{(1 + \sqrt x )}^5}} $ $ + 4\sqrt {{{(1 + \sqrt x )}^3}} – 4\sqrt {1 + \sqrt x } + C.$
d) $\int {\frac{{xdx}}{{\sqrt {{{\left( {1 + {x^2}} \right)}^3}} }}} $ $ = \int {\frac{1}{{\left( {1 + {x^2}} \right)}}} \cdot \frac{x}{{\sqrt {1 + {x^2}} }}dx.$
Đặt: $t = \sqrt {1 + {x^2}} $ $ \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{dt = \frac{x}{{\sqrt {1 + {x^2}} }}dx}\\
{{t^2} = 1 + {x^2}}
\end{array}} \right.$
$ \Rightarrow \int {\frac{{xdx}}{{\sqrt {{{\left( {1 + {x^2}} \right)}^3}} }}} $ $ = \int {\frac{{dt}}{{{t^2}}}} = – \frac{1}{t} + C$ $ = – \frac{1}{{\sqrt {1 + {x^2}} }} + C.$

Dạng 5. Nguyên hàm các hàm số lượng giác
Phương pháp
:

$u$ là hàm số theo $x$ Trường hợp đặc biệt: $u = ax + b$ $(a \ne 0)$
$\int {\cos } udu = \sin u + C$ $\int {\cos } (ax + b)dx$ $ = \frac{1}{a}\sin (ax + b) + C$
$\int {\sin } udu = – \cos u + C$ $\int {\sin } (ax + b)dx$ $ = – \frac{1}{a}\cos (ax + b) + C$
$\int {\frac{1}{{{{\cos }^2}u}}} du$ $ = \int {\left( {1 + {{\tan }^2}u} \right)} du$ $ = \tan u + C$ $\int {\frac{1}{{{{\cos }^2}(ax + b)}}} dx$ $ = \int {\left[ {1 + {{\tan }^2}(ax + b)} \right]} dx$ $ = \frac{1}{a}\tan (ax + b) + C$
$\int {\frac{1}{{{{\sin }^2}u}}} du$ $ = \int {\left( {1 + {{\cot }^2}u} \right)} du$ $ = – \cot u + C$ $\int {\frac{1}{{{{\sin }^2}(ax + b)}}} dx$ $ = \int {\left[ {1 + {{\cot }^2}(ax + b)} \right]} dx$ $ = – \frac{1}{a}\cot (ax + b) + C$

Tính $\int f (x)dx$ với $f(x) = \frac{{\alpha \sin x + \beta \cos x}}{{a\sin x + b\cos x}} = \frac{{u(x)}}{{v(x)}}$, $\left( {{\alpha ^2} + {\beta ^2} \ne 0,{a^2} + {b^2} \ne 0} \right).$
Biến đổi: $f(x) = \frac{{A.v(x) + B.v'(x)}}{{v(x)}}$ $ = \frac{{A(a\sin x + b\cos x) + B(a\cos x – b\sin x)}}{{a\sin x + b\cos x}}$ $ = \frac{{(aA – bB)\sin x + (Ab + aB)\cos x}}{{a\sin x + b\cos x}}.$
Dùng phương pháp đồng nhất thức: $(aA – bB)\sin x + (Ab + aB)\cos x$ $ = \alpha \sin x + \beta \cos x$, $\forall x \in {D_f}$ $ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{aA – bB = \alpha }\\
{Ab + aB = \beta }
\end{array}} \right.$, suy ra $A$ và $B.$
Suy ra: $\int f (x)dx$ $ = \int {\frac{{A.v(x) + B.v'(x)}}{{v(x)}}} dx$ $ = \int {Adx} + B\int {\frac{{v'(x)}}{{v(x)}}} dx$ $ = Ax + B\ln |v(x)| + C.$

Ví dụ 1. Tìm các họ nguyên hàm sau:
a) $\int {\frac{{dx}}{{{{\cos }^4}x – {{\sin }^4}x}}} .$
b) $\int {{{\sin }^4}} x.{\cos ^3}xdx.$
c) $\int {\sqrt {1 + {{\cos }^2}x} } .\sin 2xdx.$
d) $\int {{{\cos }^5}} xdx.$

a) $\int {\frac{{dx}}{{{{\cos }^4}x – {{\sin }^4}x}}} $ $ = \int {\frac{{dx}}{{{{\cos }^2}x – {{\sin }^2}x}}} $ $ = \int {\frac{{dx}}{{\cos 2x}}} $ $ = \int {\frac{{\cos 2x}}{{1 – {{\sin }^2}2x}}} dx.$
Đặt: $t = \sin 2x$ $ \Rightarrow dt = 2\cos 2xdx$ $ \Rightarrow \cos 2xdx = \frac{{dt}}{2}.$
$ \Rightarrow \int {\frac{{\cos 2x}}{{1 – {{\sin }^2}2x}}} dx$ $ = \frac{1}{2}\int {\frac{{dt}}{{1 – {t^2}}}} $ $ = – \frac{1}{2}\int {\frac{{dt}}{{(t – 1)(t + 1)}}} $ $ = – \frac{1}{4}\int {\left( {\frac{1}{{t – 1}} – \frac{1}{{t + 1}}} \right)} dt$ $ = – \frac{1}{4}\ln \left| {\frac{{t – 1}}{{t + 1}}} \right| + C$ $ = – \frac{1}{4}\ln \left| {\frac{{\sin 2x – 1}}{{\sin 2x + 1}}} \right| + C.$
b) Đặt $t = \sin x \Rightarrow dt = \cos x.dx.$
$ \Rightarrow \int {{{\sin }^4}} x{\cos ^3}xdx$ $ = \int {{{\sin }^4}} x\left( {1 – {{\sin }^2}x} \right)\cos xdx$ $ = \int {{t^4}} \left( {1 – {t^2}} \right)dt$ $ = \int {\left( {{t^4} – {t^6}} \right)} dt$ $ = \frac{{{t^5}}}{5} – \frac{{{t^7}}}{7} + C$ $ = \frac{{{{\sin }^5}x}}{5} – \frac{{{{\sin }^7}x}}{7} + C.$
c) Đặt $t = \sqrt {1 + {{\cos }^2}x} $ $ \Rightarrow {t^2} = 1 + {\cos ^2}x$ $ \Rightarrow 2 tdt = – \sin 2xdx$ $ \Rightarrow \sin 2xdx = – 2tdt.$
$ \Rightarrow \int {\sqrt {1 + {{\cos }^2}x} } .\sin 2xdx$ $ = – 2\int {{t^2}} dt = – \frac{2}{3}{t^3} + C$ $ = – \frac{2}{3}\sqrt {{{\left( {1 + {{\cos }^2}x} \right)}^3}} + C.$
d) Đặt: $t = \sin x \Rightarrow dt = \cos x.dx.$
$ \Rightarrow \int {{{\cos }^5}} xdx$ $ = \int {{{\left( {1 – {{\sin }^2}x} \right)}^2}} \cos xdx$ $ = \int {{{\left( {1 – {t^2}} \right)}^2}} dt$ $ = \int {\left( {1 – 2{t^2} + {t^4}} \right)} dt$ $ = t – \frac{2}{3}{t^3} + \frac{1}{5}{t^5} + C$ $ = \sin x – \frac{2}{3}{\sin ^3}x + \frac{1}{5}{\sin ^5}x + C.$\

Ví dụ 2. Tìm các họ nguyên hàm sau:
a) $\int {\frac{1}{{\sin x}}} dx.$
b) $\int {\frac{1}{{{{\sin }^6}x}}} dx.$
c) $\int {\frac{1}{{{{\sin }^4}x\cos x}}} dx.$
d) $\int {{{\tan }^4}} xdx.$
e) $\int {{{\tan }^6}} xdx.$
f) $\int {{{\tan }^5}} xdx.$

a)
Cách 1: $\int {\frac{1}{{\sin x}}} dx$ $ = \int {\frac{1}{{2\sin \frac{x}{2} \cdot \cos \frac{x}{2}}}} dx$ $ = \int {\frac{1}{{2\tan \frac{x}{2} \cdot {{\cos }^2}\frac{x}{2}}}} dx$ $ = \int {\frac{{d\left( {\tan \frac{x}{2}} \right)}}{{\tan \frac{x}{2}}}} $ $ = \ln \left| {\tan \frac{x}{2}} \right| + C.$
Cách 2: $\int {\frac{1}{{\sin x}}} dx$ $ = \int {\frac{{{{\sin }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2}}}{{2\sin \frac{x}{2}.\cos \frac{x}{2}}}} dx$ $ = \frac{1}{2}\int {\frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}} dx + \frac{1}{2}\int {\frac{{\cos \frac{x}{2}}}{{\sin \frac{x}{2}}}} dx$ $ = – \int {\frac{{d\left( {\cos \frac{x}{2}} \right)}}{{\cos \frac{x}{2}}}} + \int {\frac{{d\left( {\sin \frac{x}{2}} \right)}}{{\sin \frac{x}{2}}}} $ $ = – \ln \left| {\cos \frac{x}{2}} \right| + \ln \left| {\sin \frac{x}{2}} \right| + C$ $ = \ln \left| {\tan \frac{x}{2}} \right| + C.$
Cách 3: $\int {\frac{1}{{\sin x}}} dx = \int {\frac{{\sin x}}{{1 – {{\cos }^2}x}}} dx.$
Đặt: $t = \cos x$ $ \Rightarrow dt = – \sin xdx.$
$ \Rightarrow \int {\frac{1}{{\sin x}}} dx = \int {\frac{{dt}}{{{t^2} – 1}}} $ $ = \int {\frac{{dt}}{{(t – 1)(t + 1)}}} $ $ = \frac{1}{2}\int {\left( {\frac{1}{{t – 1}} – \frac{1}{{t + 1}}} \right)} dt$ $ = \frac{1}{2}\ln \left| {\frac{{t – 1}}{{t + 1}}} \right| + C$ $ = \frac{1}{2}\ln \left| {\frac{{\cos x – 1}}{{\cos x + 1}}} \right| + C$ $ = \frac{1}{2}\ln \left| {{{\tan }^2}\frac{x}{2}} \right| + C$ $ = \ln \left| {\tan \frac{x}{2}} \right| + C.$
b) $\int {\frac{1}{{{{\sin }^6}x}}} dx$ $ = \int {{{\left( {1 + {{\cot }^2}x} \right)}^2}} \cdot \frac{1}{{{{\sin }^2}x}}dx$ $ = – \int {\left( {1 + 2{{\cot }^2}x + {{\cot }^4}x} \right)} d(\cot x)$ $ = – \cot x – \frac{2}{3}{\cot ^3}x – \frac{1}{5}{\cot ^5}x + C.$
c) $\int {\frac{1}{{{{\sin }^4}x\cos x}}} dx$ $ = \int {\frac{{\cos x}}{{{{\sin }^4}x\left( {1 – {{\sin }^2}x} \right)}}} dx.$
Đặt $t = \sin x$ $ \Rightarrow dt = \cos xdx.$
$ \Rightarrow \int {\frac{1}{{{{\sin }^4}x\cos x}}} dx$ $ = \int {\frac{{dt}}{{{t^4}\left( {1 – {t^2}} \right)}}} $ $ = – \int {\left( {\frac{1}{{{t^2}}} \cdot \frac{1}{{{t^2}\left( {{t^2} – 1} \right)}}} \right)} dt$ $ = – \int {\frac{1}{{{t^2}}}} \left( {\frac{1}{{{t^2} – 1}} – \frac{1}{{{t^2}}}} \right)dt$ $ = – \int {\frac{1}{{{t^2}\left( {{t^2} – 1} \right)}}} dt + \int {\frac{1}{{{t^4}}}} dt$ $ = – \int {\frac{1}{{{t^2} – 1}}} dt + \int {\frac{1}{{{t^2}}}} dt + \int {\frac{1}{{{t^4}}}} dt$ $ = – \frac{1}{2}\int {\left( {\frac{1}{{t – 1}} – \frac{1}{{t + 1}}} \right)} dt$ $ + \int {\frac{1}{{{t^2}}}} dt + \int {\frac{1}{{{t^4}}}} dt$ $ = – \frac{1}{2}\ln \left| {\frac{{t – 1}}{{t + 1}}} \right| – \frac{1}{t} – \frac{1}{{3{t^3}}} + C$ $ = – \frac{1}{2}\ln \left| {\frac{{\sin x – 1}}{{\sin x + 1}}} \right|$ $ – \frac{1}{{\sin x}} – \frac{1}{{3{{\sin }^3}x}} + C.$
d) $\int {{{\tan }^4}} xdx$ $ = \int {\left( {{{\tan }^4}x – 1} \right)} dx + \int d x$ $ = \int {\left( {{{\tan }^2}x – 1} \right)} \left( {{{\tan }^2}x + 1} \right)dx + \int d x$ $ = \int {\left( {{{\tan }^2}x – 1} \right)} d(\tan x) + \int d x$ $ = \frac{{{{\tan }^3}x}}{3} – \tan x + x + C.$
e) $\int {{{\tan }^6}} xdx = \int {\left( {{{\tan }^6}x + 1} \right)} dx – \int d x$ $ = \int {\left( {{{\tan }^2}x + 1} \right)} \left( {{{\tan }^4}x – {{\tan }^2}x + 1} \right)dx – \int d x$ $ = \int {\left( {{{\tan }^4}x – {{\tan }^2}x + 1} \right)} d(\tan x)) – \int d x$ $ = \frac{{{{\tan }^5}x}}{5} – \frac{{{{\tan }^3}x}}{3} + \tan x – x + C.$
f) $\int {{{\tan }^5}} xdx = \int {\frac{{{{\sin }^5}x}}{{{{\cos }^5}x}}} dx$ $ = \int {\frac{{{{\sin }^4}x}}{{{{\cos }^5}x}}} \sin xdx.$
Đặt: $t = \cos x \Rightarrow dt = – \sin xdx.$
$ \Rightarrow \int {{{\tan }^5}} xdx = – \int {\frac{{{{\left( {1 – {t^2}} \right)}^2}}}{{{t^5}}}} dt$ $ = – \int {\frac{{1 – 2{t^2} + {t^4}}}{{{t^5}}}} dt$ $ = – \int {\frac{1}{{{t^5}}}} dt + 2\int {\frac{1}{{{t^3}}}} dt – \int {\frac{1}{t}dt} $ $ = \frac{1}{{4{t^4}}} – \frac{1}{{{t^2}}} – \ln |t| + C$ $ = \frac{1}{{4{{\cos }^4}x}} – \frac{1}{{{{\cos }^2}x}}$ $ – \ln |\cos x| + C.$

Ví dụ 3. Tìm các họ nguyên hàm sau:
a) $\int {\frac{{dx}}{{\sqrt 3 \cos x – \sin x}}} dx.$
b) $\int {\frac{{dx}}{{4\sin x + 3\cos x – 5}}} .$
c) $\int {\frac{{dx}}{{{{\sin }^2}x – 5\sin x.\cos x}}} .$
d) $\int {\frac{1}{{3{{\cos }^2}x + {{\sin }^2}x + 4\sin x\cos x}}} dx.$

a) $\int {\frac{{dx}}{{\sqrt 3 \cos x – \sin x}}} dx$ $ = \frac{1}{2}\int {\frac{{dx}}{{\cos \left( {x + \frac{\pi }{6}} \right)}}} $ $ = \frac{1}{2}\int {\frac{{\cos \left( {x + \frac{\pi }{6}} \right)dx}}{{1 – {{\sin }^2}\left( {x + \frac{\pi }{6}} \right)}}} .$
Đặt: $t = \sin \left( {x + \frac{\pi }{6}} \right)$ $ \Rightarrow dt = \cos \left( {x + \frac{\pi }{6}} \right)dx.$
$\int {\frac{{dx}}{{\sqrt 3 \cos x – \sin x}}} dx$ $ = – \frac{1}{2}\int {\frac{{dt}}{{{t^2} – 1}}} $ $ = – \frac{1}{4}\int {\left( {\frac{1}{{t – 1}} – \frac{1}{{t + 1}}} \right)} dt$ $ = – \frac{1}{4}\ln \left| {\frac{{t – 1}}{{t + 1}}} \right| + C$ $ = – \frac{1}{4}\ln \left| {\frac{{\sin \left( {x + \frac{\pi }{6}} \right) – 1}}{{\sin \left( {x + \frac{\pi }{6}} \right) + 1}}} \right| + C.$
b) Đặt: $t = \tan \frac{x}{2}$ $\left\{ {\begin{array}{*{20}{l}}
{dt = \frac{1}{2}\left( {1 + {{\tan }^2}\frac{x}{2}} \right)dx \Rightarrow dx = \frac{{2dt}}{{1 + {t^2}}}}\\
{\sin x = \frac{{2t}}{{1 + {t^2}}},\quad \cos x = \frac{{1 – {t^2}}}{{1 + {t^2}}}}
\end{array}} \right.$
$ \Rightarrow \int {\frac{{dx}}{{4\sin x + 3\cos x – 5}}} $ $ = \int {\frac{1}{{4\frac{{2t}}{{1 + {t^2}}} + 3\frac{{1 – {t^2}}}{{1 + {t^2}}} – 5}}} \cdot \frac{{2dt}}{{\left( {1 + {t^2}} \right)}}$ $ = – \int {\frac{{dt}}{{4{t^2} – 4t + 1}}} $ $ = – \int {\frac{{dt}}{{{{(2t – 1)}^2}}}} $ $ = \frac{1}{2} \cdot \frac{1}{{2t – 1}} + C$ $ = \frac{1}{2} \cdot \frac{1}{{2\tan \frac{x}{2} – 1}} + C.$
c) $\int {\frac{{dx}}{{{{\sin }^2}x – 5\sin x.\cos x}}} $ $ = \int {\frac{1}{{{{\cos }^2}x\left( {{{\tan }^2}x – 5\tan x} \right)}}} dx.$
Đặt: $t = \tan x$ $ \Rightarrow dt = \frac{1}{{{{\cos }^2}x}}dx.$
$ \Rightarrow \int {\frac{{dx}}{{{{\sin }^2}x – 5\sin x.\cos x}}} $ $ = \int {\frac{{dt}}{{{t^2} – 5t}}} $ $ = \frac{1}{5}\int {\left( {\frac{1}{{t – 5}} – \frac{1}{t}} \right)} dt$ $ = \frac{1}{5}\ln \left| {\frac{{t – 5}}{t}} \right| + C$ $ = \frac{1}{5}\ln \left| {\frac{{\tan x – 5}}{{\tan x}}} \right| + C.$
d) $\int {\frac{1}{{3{{\cos }^2}x + {{\sin }^2}x + 4\sin x.\cos x}}} dx$ $ = \int {\frac{1}{{{{\cos }^2}x\left( {3 + {{\tan }^2}x + 4\tan x} \right)}}} dx$ $ = \int {\frac{{d(\tan x)}}{{(\tan x + 1)(\tan x + 3)}}} $ $ = \frac{1}{2}\int {\left( {\frac{1}{{\tan x + 1}} – \frac{1}{{\tan x + 3}}} \right)} d(\tan x)$ $ = \frac{1}{2}\ln \left| {\frac{{\tan x + 1}}{{\tan x + 3}}} \right| + C.$