$\text{ Let }y=y\left(x\right)\text{ be the solution of the differential equation }{\left(x^2+1\right)}^2\frac{dy}{dx}+2x\left(x^2+1\right)y$$=1\text{ such that }\mathrm{y}\left(0\right)=0\text{ . If }8\mathrm{y}\left(1\right)=\mathrm{k}\pi \text{ then }\mathrm{k}\text{ is }$

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)\mathrm{y}=\frac{1}{{\left(1+{\mathrm{x}}^2\right)}^2}\text{ which is linear D.E }\\ \text{ I. }F=e^{\int \frac{2x}{1+x^2}dx}=e^{\log \left(1+x^2\right)}=1+x^2\\ \text{ Solution of the }D.E\text{ is }y\left(1+x^2\right)=\int \frac{1}{{\left(1+x^2\right)}^2}\cdot \left(1+x^2\right)dx=\int \frac{1}{1+x^2}dx\\ y\left(1+x^2\right)={\mathrm{Tan}}^{-1}x+c\\ \text{ Given y(0)}=0\Rightarrow \mathrm{c}=0\\ \Rightarrow \mathrm{y}\left(1+{\mathrm{x}}^2\right)={\mathrm{Tan}}^{-1}\mathrm{x}\\ \Rightarrow \mathrm{y}=\frac{{\mathrm{Tan}}^{-1}\mathrm{x}}{1+{\mathrm{x}}^2}\\ \to \mathrm{y}\left(1\right)=\frac{{\mathrm{Tan}}^{-1}\left(1\right)}{1+1}=\frac{{\displaystyle \frac{\pi }{4}}}{2}=\frac{\pi }{8}\end{array}$