Question

Evaluate $\underset{0}{\overset{1}{\int }}\frac{x^2}{1+x^2}dx$

Answer 1

We have,

$I=\underset{0}{\overset{1}{\int }}\frac{x^2}{1+x^2}dx$

$I=\underset{0}{\overset{1}{\int }}\frac{x^2+1-1}{1+x^2}dx$

$I={\int }_0^{1}1dx-\underset{0}{\overset{1}{\int }}\frac{1}{1+x^2}dx$

$I=[x{]}_0^1-[{\tan }^{-1}x{]}_0^1$

$I=[1-0]-({\tan }^{-1}1-{\tan }^{-1}0)$

$I=1-{\tan }^{-1}\tan \frac{\pi }{4}+0$

$I=1-\frac{\pi }{4}$

Hence, this is the answer.