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 $u=1-x$, $dv=\cos xdx$

=> $du=-dx$

     $v=\sin x$

=> $\int (1-x)\cos xdx=(1-x)\sin x+\int \sin xdx$

<=> $\int (1-x)\cos xdx=(1-x)\sin x-\cos x+C$