$\int_{0}^{1}\left(\frac{x^{2}-2}{x^{2}+1}\right) d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} \int_{0}^{1} \frac{x^{2}+1-3}{x^{2}+1} d x &=\int_{0}^{1} \frac{x^{2}+1}{x^{2}+1}-\frac{3}{x^{2}+1} d x \ &=\int_{0}^{1} 1-\frac{3}{x^{2}+1} d x=\left[x-3 	an ^{-1} xight]_{0}^{1} \ &=\left(1-3 	an ^{-1} 1ight)-\left(0-3 	an ^{-1} 0ight)=1- \frac{3 \pi}{4} \end{aligned}$

$\int_{0}^{1}\left(\frac{x^{2}-2}{x^{2}+1}\right) d x=$

Solution