Sử dụng phương pháp nguyên hàm từng phần để tìm nguyên hàm

Công thức đúng là $\int {udv}  = uv - \int {vdu}$.

Ta có: $u = g\left( x \right) \Rightarrow du = g'\left( x \right)dx$.

$dv = h\left( x \right)dx \Rightarrow v = \int {h\left( x \right)dx}$.

Đặt $\left\{ \begin{array}{l}u = x + 1\\dv = f'\left( x \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = f\left( x \right)\end{array} \right.$  

$\Rightarrow F\left( x \right) = \left( {x + 1} \right)f\left( x \right) - \int {f\left( x \right)dx}  + C$

$\Rightarrow I = \int {f\left( x \right)dx}  = \left( {x + 1} \right)f\left( x \right) - F\left( x \right) + C.$

Đặt $\left\{ \begin{array}{l}u = \ln 3x\\dv = {x^2}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{3}{{3x}}dx\\v = \dfrac{1}{3}{x^3}\end{array} \right.$

$\Rightarrow I = \dfrac{1}{3}{x^3}\ln 3x - \int {\dfrac{1}{3}{x^3}.\dfrac{3}{{3x}}dx}  = \dfrac{1}{3}{x^3}\ln 3x - \int {\dfrac{1}{3}{x^2}dx}  = \dfrac{1}{3}{x^3}\ln 3x - \dfrac{1}{9}{x^3} + C$

Đặt $\left\{ \begin{array}{l}u = \ln 3x\\dv = {x^3}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{x}dx\\v = \dfrac{{{x^4}}}{4}\end{array} \right.$ 

$\Rightarrow I = \dfrac{1}{4}{x^4}\ln 3x - \dfrac{1}{4}\int {{x^3}dx}  + C = \dfrac{1}{4}{x^4}\ln 3x - \dfrac{{{x^4}}}{{16}} + C$

Ta có: $f'(x) = \left( {x + 1} \right){e^x} \Rightarrow f\left( x \right) = \int {\left( {x + 1} \right){e^x}dx}$.

Đặt: $\left\{ \begin{array}{l}u = x + 1\\dv = {e^x}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = {e^x}\end{array} \right.$

$\Rightarrow I = \left( {x + 1} \right){e^x} - \int {{e^x}} dx = x{e^x} + {e^x} - {e^x} + C = x{e^x} + C$

Do đó ta được $a = 1;b = 0 \Rightarrow a + b = 1$.

Đặt $\left\{ \begin{array}{l}u = 2x + 3\\dv = {e^x}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = 2dx\\v = {e^x}\end{array} \right.$.$\Rightarrow \int {(2x + 3){e^x}dx}  = (2x + 3){e^x} - \int {{e^x}2dx = } (2x + 3){e^x} - 2{e^x} = (2x + 1){e^x}$

Khi đó $a = 2,b = 1$

$F\left( x \right) = \int {\dfrac{x}{{{e^x}}}dx}  = \int {x{e^{ - x}}dx}$

Đặt $\left\{ \begin{array}{l}u = x\\dv = {e^{ - x}}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v =  - {e^{ - x}}\end{array} \right. \Rightarrow F\left( x \right) =  - x{e^{ - x}} + \int {{e^{ - x}}dx}  + C \\ =  - x{e^{ - x}} - {e^{ - x}} + C =  - \left( {x + 1} \right){e^{ - x}} + C =  - \dfrac{{x + 1}}{{{e^x}}} + C.$

$- \dfrac{{x + a}}{{{e^x}}}$ là một họ nguyên hàm của hàm số $f\left( x \right) = \dfrac{x}{{{e^x}}} \Rightarrow \left\{ \begin{array}{l}a = 1\\C = 0\end{array} \right..$

$F\left( x \right) = \int {\dfrac{{{2^x} - 1}}{{{e^x}}}dx}  = \int {\left( {{2^x} - 1} \right){e^{ - x}}dx}  = \int {{2^x}{e^{ - x}}dx}  - \int {{e^{ - x}}dx}$

$= \int {{2^x}{e^{ - x}}dx}  + {e^{ - x}} + {C_1} = I + {e^{ - x}} + {C_1}.$

Đặt $\left\{ \begin{array}{l}u = {2^x}\\dv = {e^{ - x}}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = {2^x}\ln 2dx\\v =  - {e^{ - x}}\end{array} \right.$  

$\begin{array}{l} \Rightarrow I =  - {2^x}{e^{ - x}} + \ln 2\int {{2^x}{e^{ - x}}dx}  + {C_2} =  - {2^x}{e^{ - x}} + \ln 2.I + {C_2} \\ \Leftrightarrow \left( {\ln 2 - 1} \right)I + {C_2} = {2^x}{e^{ - x}} \Rightarrow I = \dfrac{{{2^x}{e^{ - x}}}}{{\ln 2 - 1}} + {C_2}.\\ \Rightarrow F\left( x \right) = \dfrac{{{2^x}{e^{ - x}}}}{{\ln 2 - 1}} + {e^{ - x}} + C = \dfrac{{{2^x}}}{{\left( {\ln 2 - 1} \right){e^x}}} + \dfrac{1}{{{e^x}}} + C\\ \Rightarrow F\left( 0 \right) = \dfrac{1}{{\ln 2 - 1}} + 1 + C = 1 \Rightarrow C =  - \dfrac{1}{{\ln 2 - 1}}\\ \Rightarrow F\left( x \right) = \dfrac{{{2^x}}}{{\left( {\ln 2 - 1} \right){e^x}}} + \dfrac{1}{{{e^x}}} - \dfrac{1}{{\ln 2 - 1}} = \dfrac{1}{{\ln 2 - 1}}{\left( {\dfrac{2}{e}} \right)^x} + {\left( {\dfrac{1}{e}} \right)^x} - \dfrac{1}{{\ln 2 - 1}}.\end{array}$

$I = \int {x\sin x\cos xdx}  = \dfrac{1}{2}\int {x\sin 2xdx}$

Đặt $\left\{ \begin{array}{l}u = x\\dv = \sin 2xdx\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}du = dx\\v =  - \dfrac{{\cos 2x}}{2}\end{array} \right.$

$\Rightarrow I = \dfrac{1}{2}\left( { - x.\dfrac{{\cos 2x}}{2} + \dfrac{1}{2}\int {\cos 2xdx} } \right) + C$

$= \dfrac{1}{2}\left( { - \dfrac{{x\cos 2x}}{2} + \dfrac{{\sin 2x}}{4}} \right) + C$

Đặt $\sqrt x  = t \Rightarrow x = {t^2} \Rightarrow dx = 2tdt \Rightarrow I = 2\int {t\cos tdt} .$

Đặt $\left\{ \begin{array}{l}u = t\\dv = \cos tdt\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dt\\v = \sin t\end{array} \right.$

$\Rightarrow I = 2\left( {t\sin t - \int {{\mathop{\rm sint}\nolimits} dt}  + C} \right) = 2\left( {t\sin t + \cos t + C} \right)$

$= 2\left( {\sqrt x \sin \sqrt x  + \cos \sqrt x } \right) + C.$

Ta có $F\left( x \right) = \int {x.\cos xdx}$

Đặt $\left\{ \begin{array}{l}u = x\\dv = \cos xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = \sin x\end{array} \right. \Rightarrow F\left( x \right) = x\sin x - \int {\sin xdx}  + C = x\sin x + \cos x + C.$ $F\left( 0 \right) = 1 \Leftrightarrow 0\sin 0 + \cos 0 + C = 1 \Leftrightarrow 1 + C = 1 \Leftrightarrow C = 0 \Rightarrow F\left( x \right) = x\sin x + \cos x$

Ta có: $F\left( { - x} \right) = \left( { - x} \right)\sin \left( { - x} \right) + \cos \left( { - x} \right) = x\sin x + \cos x = F\left( x \right) \Rightarrow F\left( x \right)$ là hàm chẵn.

$\begin{array}{l}\left\{ \begin{array}{l}u = x\\dv = \dfrac{1}{{{{\cos }^2}x}}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = \tan x\end{array} \right.\\ \Rightarrow F\left( x \right) = x\tan x - \int {\tan xdx}  + C = x\tan x - \int {\dfrac{{\sin x}}{{\cos x}}dx}  + C \\ = x\tan x + \int {\dfrac{{d\left( {\cos x} \right)}}{{\cos x}}}  + C = x\tan x + \ln \left| {\cos x} \right| + C.\\ \Rightarrow F\left( 0 \right) = C = 0 \Rightarrow F\left( \pi  \right) = 0\end{array}$

Vì $x{e^x}$ là một nguyên hàm của hàm số $f\left( { - x} \right)$ nên $\left( {x{e^x}} \right)' = f\left( { - x} \right) \Leftrightarrow f\left( { - x} \right) = {e^x} + x{e^x} = {e^x}\left( {1 + x} \right)$.

$\Rightarrow f\left( x \right) = {e^{ - x}}\left( {1 - x} \right)$.

$\begin{array}{l} \Rightarrow f'\left( x \right) =  - {e^{ - x}}\left( {1 - x} \right) - {e^{ - x}} =  - {e^{ - x}}\left( {2 - x} \right) = \left( {x - 2} \right){e^{ - x}}\\ \Rightarrow f'\left( x \right){e^x} = \left( {x - 2} \right){e^{ - x}}.{e^x} = x - 2\\ \Rightarrow F\left( x \right) = \int\limits_{}^{} {f\left( x \right)dx}  = \int\limits_{}^{} {\left( {x - 2} \right)dx}  = \dfrac{{{x^2}}}{2} - 2x + C\\F\left( 0 \right) = 1 \Rightarrow C = 1 \Rightarrow F\left( x \right) = \dfrac{{{x^2}}}{2} - 2x + 1\\ \Rightarrow F\left( { - 1} \right) = \dfrac{{{{\left( { - 1} \right)}^2}}}{2} - 2\left( { - 1} \right) + 1 = \dfrac{7}{2}\end{array}$

$I = \int {x{{\tan }^2}xdx}  = \int {x\left( {\dfrac{1}{{{{\cos }^2}x}} - 1} \right)dx}  = \int {x.\dfrac{1}{{{{\cos }^2}x}}dx}  - \int {xdx}  = {I_1} - {I_2}$

Ta có: ${I_2} = \int {xdx}  = \dfrac{{{x^2}}}{2} + {C_2},{I_1} = \int {x\dfrac{1}{{{{\cos }^2}x}}dx}$

Đặt $\left\{ \begin{array}{l}u = x\\dv = \dfrac{1}{{{{\cos }^2}x}}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = \tan x\end{array} \right.$

$\begin{array}{l} \Rightarrow {I_1} = x\tan x - \int {\tan xdx}  + {C_1} = x\tan x - \int {\dfrac{{\sin x}}{{\cos x}}dx}  + {C_1} \\ = x\tan x + \int {\dfrac{{d\left( {\cos x} \right)}}{{\cos x}}}  + {C_1} = x\tan x + \ln \left| {\cos x} \right| + {C_1}.\\ \Rightarrow I = x\tan x + \ln \left| {\cos x} \right| + {C_1} - \dfrac{{{x^2}}}{2} - {C_2} = x\tan x + \ln \left| {\cos x} \right| - \dfrac{{{x^2}}}{2} + C.\end{array}$

Ta có:

$\begin{array}{l}\cos 2x\ln \left( {\sin x + \cos x} \right) = \left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)\ln \left( {\sin x + \cos x} \right)\\ \Rightarrow I = \int {\left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right)\ln \left( {\sin x + \cos x} \right)dx} \end{array}$

Đặt $t = \sin x + \cos x \Rightarrow dt = \left( {\cos x - \sin x} \right)dx$ , khi đó ta có:$I = \int {t\ln tdt}$

Đặt $\left\{ \begin{array}{l}u = \ln t\\dv = tdt\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{t}dt\\v = \dfrac{{{t^2}}}{2}\end{array} \right.$

$\begin{array}{l} \Rightarrow I = \dfrac{1}{2}{t^2}\ln t - \dfrac{1}{2}\int {tdt}  + C = \dfrac{1}{2}{t^2}\ln t - \dfrac{{{t^2}}}{4} + {C_1}\\\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}{\left( {\sin x + \cos x} \right)^2}\ln \left( {\sin x + \cos x} \right) - \dfrac{{{{\left( {\sin x + \cos x} \right)}^2}}}{4} + {C_1}\\\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}\left( {{{\sin }^2}x + {{\cos }^2}x + \sin 2x} \right)\ln \left( {\sin x + \cos x} \right) - \dfrac{{1 + \sin 2x}}{4} + {C_1}\\\,\,\,\,\,\,\,\,\, = \dfrac{1}{4}\left( {1 + \sin 2x} \right)\ln {\left( {\sin x + \cos x} \right)^2} - \dfrac{{\sin 2x}}{4} - \dfrac{1}{4} + {C_1}\\\,\,\,\,\,\,\,\,\, = \dfrac{1}{4}\left( {1 + \sin 2x} \right)\ln \left( {1 + \sin 2x} \right) - \dfrac{{\sin 2x}}{4} + C.\end{array}$  

Đặt $\left\{ \begin{array}{l}u = \ln \left( {x + \sqrt {{x^2} + 1} } \right)\\dv = dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{{1 + \dfrac{x}{{\sqrt {{x^2} + 1} }}}}{{x + \sqrt {{x^2} + 1} }}dx\\v = x\end{array} \right.$  

$\Rightarrow \left\{ \begin{array}{l}du = \dfrac{{\dfrac{{x + \sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + 1} }}}}{{x + \sqrt {{x^2} + 1} }}dx = \dfrac{{dx}}{{\sqrt {{x^2} + 1} }}\\v = x\end{array} \right.$

$\Rightarrow I = x\ln \left( {x + \sqrt {{x^2} + 1} } \right) - \int {\dfrac{x}{{\sqrt {{x^2} + 1} }}dx}  + {C_1}.$

Đặt $t = \sqrt {{x^2} + 1}  \Rightarrow {t^2} = {x^2} + 1 \Leftrightarrow tdt = xdx$

$\Rightarrow \int {\dfrac{x}{{\sqrt {{x^2} + 1} }}dx}  = \int {\dfrac{{tdt}}{t}}  = \int {dt}  = t + {C_2} = \sqrt {{x^2} + 1}  + {C_2}$

Khi đó ta có:  $\Rightarrow I = x\ln \left( {x + \sqrt {{x^2} + 1} } \right) - \sqrt {{x^2} + 1}  + C.$

Đặt $\left\{ \begin{array}{l}u = {e^{2x}}\\dv = \cos 3xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = 2{e^{2x}}dx\\v = \dfrac{{\sin 3x}}{3}\end{array} \right. \Rightarrow I = \dfrac{1}{3}{e^{2x}}\sin 3x - \dfrac{2}{3}\int {{e^{2x}}\sin 3xdx}  + {C_1}.$

Xét nguyên hàm $\int {{e^{2x}}\sin 3xdx}$, đặt

$\left\{ \begin{array}{l}a = {e^{2x}}\\db = \sin 3xdx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}da = 2{e^{2x}}\\b =  - \dfrac{{\cos 3x}}{3}\end{array} \right.$

$\Rightarrow \int {{e^{2x}}\sin 3xdx}  =  - \dfrac{1}{3}{e^{2x}}\cos 3x + \dfrac{2}{3}\int {{e^{2x}}\cos 3x}  + {C_1} =  - \dfrac{1}{3}{e^{2x}}\cos 3x + \dfrac{2}{3}I + {C_2}$

Do đó ta có 

$\begin{array}{l}I = \dfrac{1}{3}{e^{2x}}\sin 3x - \dfrac{2}{3}\left( { - \dfrac{1}{3}{e^{2x}}\cos 3x + \dfrac{2}{3}I + {C_2}} \right) + {C_1}\\ \Leftrightarrow \dfrac{{13}}{9}I = \dfrac{1}{3}{e^{2x}}\sin 3x + \dfrac{2}{9}{e^{2x}}\cos 3x + C\\ \Leftrightarrow I = \dfrac{1}{{13}}{e^{2x}}\left( {3\sin 3x + 2\cos 3x} \right) + C.\end{array}$

Ta có: $I = \int {\dfrac{{\left( {{x^2} + x} \right){e^x}}}{{x + {e^{ - x}}}}dx}  = \int {\dfrac{{\left( {{x^2} + x} \right){e^x}}}{{\dfrac{{x{e^x} + 1}}{{{e^x}}}}}dx}  = \int {\dfrac{{\left( {{x^2} + x} \right){e^{2x}}}}{{x{e^x} + 1}}dx}  = \int {\dfrac{{x{e^x}\left( {x + 1} \right){e^x}}}{{x{e^x} + 1}}dx} .$

Đặt $\left\{ \begin{array}{l}u = x{e^x}\\dv = \dfrac{{\left( {x + 1} \right){e^x}}}{{x{e^x} + 1}}dx = \dfrac{{d\left( {x{e^x} + 1} \right)}}{{x{e^x} + 1}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \left( {{e^x} + x{e^x}} \right)dx = \left( {x + 1} \right){e^x}dx\\v = \ln \left| {x{e^x} + 1} \right|\end{array} \right.$

Khi đó ta có: $I = x{e^x}\ln \left| {x{e^x} + 1} \right| - \int {\ln \left| {x{e^x} + 1} \right|\left( {x + 1} \right){e^x}dx}  + C.$ 

Đặt $t = x{e^x} + 1 \Rightarrow dt = \left( {{e^x} + x{e^x}} \right)dx = \left( {x + 1} \right){e^x}dx$

$\Rightarrow \int {\ln \left| {x{e^x} + 1} \right|\left( {x + 1} \right){e^x}dx}  = \int {\ln \left| t \right|dt}$

Đặt $\left\{ \begin{array}{l}u = \ln \left| t \right|\\dv = dt\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \dfrac{1}{t}dt\\v = t\end{array} \right.$

$\Rightarrow \int {\ln \left| t \right|dt}  = \ln \left| t \right|.t - \int {dt}  + C = \ln \left| t \right|.t - t + C$

$= \left( {x{e^x} + 1} \right)\ln \left| {x{e^x} + 1} \right| - \left( {x{e^x} + 1} \right) + C.$

Vậy $I = x{e^x}\ln \left| {x{e^x} + 1} \right| - \left( {x{e^x} + 1} \right)\ln \left| {x{e^x} + 1} \right| + \left( {x{e^x} + 1} \right) + C$

$= x{e^x} + 1 - \ln \left| {x{e^x} + 1} \right| + C.$

Ta có: $\dfrac{{{x^2} - 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} = \dfrac{{2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}} - \dfrac{1}{{{x^2} + 1}}$

$\Rightarrow \int {\dfrac{{{x^2} - 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx}  = \int {\dfrac{{2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx}  - \int {\dfrac{1}{{{x^2} + 1}}dx} \,\,\left( 1 \right)$

Ta tính $\int {\dfrac{{2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx}  = \int {\dfrac{{xd\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}}$ bằng phương pháp tích phân từng phân như sau:

Đặt $\left\{ \begin{array}{l}u = x\\dv = \dfrac{{d\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v =  - \dfrac{1}{{{x^2} + 1}}\end{array} \right.$

$\Rightarrow \int {\dfrac{{xd\left( {{x^2} + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}}  =  - \dfrac{x}{{{x^2} + 1}} + \int {\dfrac{{dx}}{{{x^2} + 1}}}  + C\,\,\left( 2 \right)$

Từ (1) và (2) suy ra  $\int {\dfrac{{{x^2} - 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx}  =  - \dfrac{x}{{{x^2} + 1}} + \int {\dfrac{{dx}}{{{x^2} + 1}}}  + C - \int {\dfrac{1}{{{x^2} + 1}}dx}  =  - \dfrac{x}{{{x^2} + 1}} + C.$