Question

Solve:$\int xlog(1+x^2)dx$

Answer 1

$\int \underset{II}{x}\underset{I}{log(1+x^2)}dx$

$=log(1+x^2)\int x\cdot dx-\int {\left[log\right(1+x^2\left)\right]}^{\prime }\int xdx\cdot dx$

$=\frac{x^2}{2}\cdot log(1+x^2)-\int \frac{2x}{1+x^2}\times \frac{x^2}{2}\cdot dx$

$=\frac{x^2}{2}\cdot log(1+x^2)-\int \underset{I_1}{\frac{x^3}{1+x^2}}\cdot dx$ $I_1:1+x^2\left)x^3\right(x$

$=\frac{x^2}{2}\cdot log(1+x^2)-\int \frac{x^3}{1+x^2}\cdot dx$ $x^3+x$

$=\frac{x^2}{2}\cdot log(1+x^2)-\frac{x^2}{2}+\frac{1}{2}\int \frac{2x}{1+x^2}\cdot dx$ _________

Let $(1+x^2)=t$ $-x$

$2x\cdot dx=dt$

$=\frac{x^2}{2}\cdot log(1+x^2)-\frac{x^2}{2}+\frac{1}{2}\int \frac{dt}{t}$

$=\frac{x^2}{2}log(1+x^2)-\frac{x^2}{2}+\frac{1}{2}logt+c$

$=\frac{x^2}{2}log(1+x^2)-\frac{x^2}{2}+\frac{1}{2}log|1+x^2|+c$.