Question

Integrate
${\int }_0^1\frac{x^3}{x^2+1}dx$

Answer 1

Consider the given integral.

$I={\int }_0^1\frac{x^3}{x^2+1}dx$

$I={\int }_0^1(x-\frac{x}{x^2+1})dx$

$I={\int }_0^1(x-\frac{2x}{2(x^2+1)})dx$

$I={\left[\frac{x^2}{2}\right]}_0^1-\frac{1}{2}{\left[{\log }_e(x^2+1)\right]}_0^1$

$I=[\frac{1^2}{2}-0]-\frac{1}{2}[{\log }_e(1^2+1)-{\log }_e(0+1)]$

$I=\frac{1}{2}-\frac{1}{2}[{\log }_e\left(2\right)-{\log }_e\left(1\right)]$

$I=\frac{1}{2}-\frac{1}{2}[{\log }_e\left(2\right)-0]$

$I=\frac{1}{2}-\frac{1}{2}{\log }_{e}2$

Hence, this is the answer.