# Q. 2# (x+1)frac{dy}{dx}=x(y^{2}+1)

Q. 2# $\left(x+1\right)\frac{dy}{dx}=x\left({y}^{2}+1\right)$
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Here the Differential equations is given by
$\left(x+1\right)\frac{dy}{dx}=x\left({y}^{2}+1\right)$
$\to \frac{dy}{{y}^{2}+1}=\frac{xdx}{x+1}$
$\to \int \frac{dy}{{y}^{2}+1}=\int \frac{xdx}{x+1}$
$\to {\mathrm{tan}}^{-1}\left(y\right)=\int \frac{x+1-1dx}{\left(x+1\right)+c}$
$\to {\mathrm{tan}}^{-1}\left(y\right)=\int \frac{\left(x+1-1\right)dx\right)}{x+1+c}$
$\to {\mathrm{tan}}^{-1}\left(y\right)=\int dx-\int \frac{dx}{\left(x+1\right)+c}$
$\to {\mathrm{tan}}^{-1}\left(y\right)=x-\mathrm{ln}\mid x+1\mid +c$
$\to y=tan\left(x-\mathrm{ln}\mid x+1\mid +c\right),$
where cc is a arbitrary constant.