$\int_{0}^{1} \frac{x^{2}}{1+x^{2}} d x=$

Question

Aicharya · Accepted Answer

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

$\int_{0}^{1} \frac{x^{2}}{1+x^{2}} d x=$

Solution