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Answer: $\tan ^{-1} x+\tan ^{-1} y+\frac{1}{2} \log \left|1+x^{2}\right|+\frac{1}{2} \log \left|1+y^{2}\right|=c$

Hint: Separate the terms of x and y and then integrate them.

Given: $(1+x)\left(1+y^{2}\right) d x+(1+y)\left(1+x^{2}\right) d y=0$

Solution:

\begin{aligned} &(1+x)\left(1+y^{2}\right) d x+(1+y)\left(1+x^{2}\right) d y=0 \\\\ &(1+x)\left(1+y^{2}\right) d x=-(1+y)\left(1+x^{2}\right) d y \\\\ &\frac{(1+x)}{1+x^{2}} d x=-\frac{(1+y)}{1+y^{2}} d y \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{(1+x)}{1+x^{2}} d x=-\int \frac{(1+y)}{1+y^{2}} d y \\\\ &\Rightarrow \int \frac{1}{1+x^{2}} d x+\int \frac{x}{1+x^{2}} d x=-\int \frac{1}{1+y^{2}} d y-\int \frac{y}{1+y^{2}} d y \end{aligned}

\begin{aligned} &1+x^{2}=t \Rightarrow 2 x d x=d t \\\\ &1+y^{2}=u \Rightarrow 2 y d y=d u \\\\ &\int \frac{1}{1+x^{2}} d x+\frac{1}{2} \int \frac{1}{t} d t=-\int \frac{1}{1+y^{2}} d y-\frac{1}{2} \int \frac{1}{u} d u \end{aligned}

\begin{aligned} &\tan ^{-1} x+\frac{1}{2} \log |t|=-\tan ^{-1} y-\frac{1}{2} \log |u|+c \\\\ &\tan ^{-1} x+\frac{1}{2} \log \left|1+x^{2}\right|=-\tan ^{-1} y-\frac{1}{2} \log \left|1+y^{2}\right|+c \\\\ &\tan ^{-1} x+\tan ^{-1} y+\frac{1}{2} \log \left|1+x^{2}\right|+\frac{1}{2} \log \left|1+y^{2}\right|=c \end{aligned}

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