Giải câu 4 bài: Nguyên hàm.
a) $\int x\ln (1+x)dx$
Đặt $u= \ln(1+x)$ , $dv= xdx$
=> $du=\frac{1}{1-x}dx$ ,
$v=\frac{x^{2}}{2}
Ta có: $\int x\ln(1+x)dx = \frac{x^{2}}{2}\ln(1+x)−\int \frac{x^{2}dx}{2(x+1)}$
<=> $\int x\ln(1+x)dx= $\frac{1}{2}(x^{2}-1)\ln(1-x)-\frac{x^{2}}{4}+\frac{x}{2}+C$
b) $\int (x^{2}+2x-1)e^{x}dx$
Đặt $u(x)=x+1$, $e^{x}dx=dv$
=> $du=dx$
$v=e^{x}$
=> $\int (x^{2}+2x-1)e^{x}dx=(x^{2}+2x-1)e^{x}-2e^{x}+C= $(x^{2}-1)e^{x}+C$
c) $\int x\sin x(2x+1)dx$
Đặt $u=x$, $dv=\sin (2x+1)dx$
=> $du=dx$
$v=-\frac{1}{2}\cos(2x+1)$
=> $\int x\sin x(2x+1)dx=-\frac{1}{2}x\cos(2x+1)+\frac{1}{2}\int \cos(2x+1)dx$
<=> $\int x\sin x(2x+1)dx=-\frac{1}{2}x\cos(2x+1)+\frac{1}{4}\sin(2x+1)+C$
d) $\int (1-x)\cos xdx$
Đặt $u=1-x$, $dv=\cos xdx$
=> $du=-dx$
$v=\sin x$
=> $\int (1-x)\cos xdx=(1-x)\sin x+\int \sin xdx$
<=> $\int (1-x)\cos xdx=(1-x)\sin x-\cos x+C$