Giải câu 3 bài: Nguyên hàm

a) $\int (1-x)^{9}dx$  

Đặt $u=1-x  =>du=-dx$

=>  $\int (1-x)^{9}dx=-\int u^{9du}=\frac{-u^{10}}{10}+C$  

<=> $\int (1-x)^{9}dx=-\int u^{9du}=\frac{-(1-x)^{10}}{10}+C$  

b) $\int x(1+x^{2})^{\frac{3}{2}}dx$  

Đặt $u=1+x^{2} => du=2dx$

=> $\int x(1+x^{2})^{\frac{3}{2}}dx=\frac{1}{2}u^{\frac{3}{2}}du$

<=> $\int x(1+x^{2})^{\frac{3}{2}}dx=\frac{\frac{1}{2}u^{\frac{3}{2}+1}}{\frac{3}{2}+1}+C$

<=> $\int x(1+x^{2})^{\frac{3}{2}}dx=\frac{(1+x^{2})^{\frac{5}{2}+1}}{5}+C$

c) $\int \cos ^{3}x\sin xdx$  

Đặt t=\cos x => dt=-\sin xdx$

=> $\int \cos ^{3}x\sin xdx=-\int t^{3}dt$  

<=> $\int \cos ^{3}x\sin xdx=-\frac{t^{4}}{4}+C$

<=> $\int \cos ^{3}x\sin xdx=-\frac{\cos^{4}x}{4}+C$

d) $\int \frac{dx}{e^{x}+e^{-x}+2}$  

Đặt $u=e^{x}+1$

=> $\int \frac{dx}{e^{x}+e^{-x}+2}=\int \frac{du}{u^{2}}$ 

<=> $\int \frac{dx}{e^{x}+e^{-x}+2}=-\frac{1}{u}+C$ 

<=> $\int \frac{dx}{e^{x}+e^{-x}+2}=-\frac{1}{e^{x}+1}+C$