Giải bài 3 trang 101 – SGK môn Giải tích lớp 12

Hướng dẫn: 

Nếu $\int{f\left( u \right)du}=F\left( u \right)+C$ và $u=u\left( x \right)$ là hàm số có đạo hàm liên tục thì $\int{f\left( u\left( x \right) \right)u'\left( x \right)dx}=F\left( u\left( x \right) \right)+C$
 

a) $\int{{{\left( 1-x \right)}^{9}}\,dx}$          

Đặt $u=1-x\Rightarrow du=-dx$

$\begin{align}\int{{{\left( 1-x \right)}^{9}}\,dx}& =-\int{{{u}^{9}}du}\\&=-\dfrac{{{u}^{10}}}{10}+C \\&=-\dfrac{{{\left( 1-x \right)}^{10}}}{10}+C\end{align}$                                

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

Đặt $u=1+{{x}^{2}}\Rightarrow du=2xdx\Rightarrow xdx=\dfrac{du}{2}$

$\begin{align} \int{x\left( 1+{{x}^{2}} \right)\,dx}&=\int{\dfrac{u}{2}du}\\&=\dfrac{{{u}^{2}}}{4}+C\\&=\dfrac{{{\left( 1+{{x}^{2}} \right)}^{2}}}{4}+C\end{align}$

c) $\int{{{\cos }^{3}}x\sin xdx}$     

Đặt $t=\cos x\Rightarrow dt=-\sin xdx$     

$\begin{align} \int{{{\cos }^{3}}x\sin xdx}&=-\int{{{t}^{3}}dt}\\&=-\dfrac{{{t}^{4}}}{4}+C\\&=-\dfrac{{{\cos }^{4}}x}{4}+C \end{align}$                               

d) $\int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}+2}}$ 

Đặt $u={{e}^{x}}+1\Rightarrow du={{e}^{x}}dx$

$\begin{align} \int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}+2}}&=\int{\dfrac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}dx}\\& =\int{\dfrac{du}{{u}^{2}}}\\& =-\dfrac{1}{u}+C\\&=-\dfrac{1}{{{e}^{x}}+1}+C \end{align}$