Question

Integrate the function.
$\int (x^2+1)logxdx.$

Answer 1

Let $I=\int (x^2+1)logxdx$
On taking log x as first function and $(x^2+1)$ as second function and integrating by parts, we get
$I=logx\int (x^2+1)dx-\int \left[\frac{d}{dx}\right(logx)\int (x^2+1\left)dx\right]\Rightarrow I=logx(\frac{x^3}{3}+x)-\int \frac{1}{x}.(\frac{x^3}{3}+x)dx\Rightarrow I=(\frac{x^3}{3}+x)logx-\int (\frac{x^2}{3}+1)dx=(\frac{x^3}{3}+x)logx-\frac{x^3}{9}-x+C$