Giải bài 4 trang 100 – SGK môn Giải tích lớp 12

Hướng dẫn: 

Nếu hai hàm số $u=u\left( x \right)$ và $v=v\left( x \right)$ có đạo hàm liên tục trên K thì $\int{u\left( x \right)v'\left( x \right)dx}=u\left( x \right)v\left( x \right)-\int{u'\left( x \right)v\left( x \right)dx}$.

a) $\int{x\ln \left( 1+x \right)dx};$

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

$\begin{aligned} \int{x\ln \left( 1+x \right)dx}&=\dfrac{{{x}^{2}}}{2}\ln \left( 1+x \right)-\int{\dfrac{{{x}^{2}}}{2}.\dfrac{1}{x+1}}dx \\ & =\dfrac{{{x}^{2}}}{2}\ln \left( 1+x \right)-\dfrac{1}{2}\int{\left( x-1+\dfrac{1}{x+1} \right)dx} \\ & =\dfrac{{{x}^{2}}}{2}\ln \left( 1+x \right)-\dfrac{{{x}^{2}}}{4}+\dfrac{x}{2}-\dfrac{1}{2}\ln \left| x+1 \right|+C \\ \end{aligned}$

b) $\int{\left( {{x}^{2}}+2x-1 \right){{e}^{x}}dx};$

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

$\begin{aligned}\int{\left( {{x}^{2}}+2x-1 \right){{e}^{x}}dx}&=\left( {{x}^{2}}+2x-1 \right){{e}^{x}}-2\int{\left( x+1 \right){{e}^{x}}dx} \\ \end{aligned}$

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

$\begin{aligned}\Rightarrow \int{\left( x+1 \right){{e}^{x}}dx}&=\left( x+1 \right){{e}^{x}}-\int{{{e}^{x}}dx} \\ & =x{{e}^{x}}+C \\ \end{aligned}\\ \begin{aligned} \Rightarrow \int{\left( {{x}^{2}}+2x-1 \right){{e}^{x}}dx}&=\left( {{x}^{2}}+2x-1 \right){{e}^{x}}-2x{{e}^{x}}+C \\ & =\left( {{x}^{2}}-1 \right){{e}^{x}}+C \\ \end{aligned}$

c) $\int{x\sin \left( 2x+1 \right)dx};$

Đặt $\left\{ \begin{aligned} & x=u \\ & \sin \left( 2x+1 \right)dx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=-\dfrac{\cos \left( 2x+1 \right)}{2} \\ \end{aligned} \right.$

$\begin{aligned}\int{x\sin \left( 2x+1 \right)dx}&=-\dfrac{x\cos \left( 2x+1 \right)}{2}+\int{\dfrac{\cos \left( 2x+1 \right)}{2}dx} \\ & =-\dfrac{x\cos \left( 2x+1 \right)}{2}+\dfrac{\sin \left( 2x+1 \right)}{4}+C \end{aligned}$

d) $\int{\left( 1-x \right)\cos xdx.}$

Đặt $\left\{ \begin{aligned} & 1-x=u \\ & \cos xdx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=-dx \\ & v=\sin x \\ \end{aligned} \right.$

$\begin{aligned}\int{\left( 1-x \right)\cos xdx}&=\left( 1-x \right)\sin x+\int{\sin xdx} \\ & =\left( 1-x \right)\sin x-\cos x+C \end{aligned}$