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