Given $(1+y^2)\tan^{-1}xdx+2y(1+x^2)dy=0$
We can rewrite this as
$2y(1+x^2)dy=-(1+y^2)\tan ^{-1} x dx$
$2 \large\frac{dy}{dx}=\frac{-(1+y^2)\tan ^{-1}x}{(1+x^2)}$
Now seperating the variables we get
$2 \large\frac{dy}{1+y^2}=\frac{-\tan ^{-1}(x)}{1+x^2}$$dx$
Now integrating we get,
$\int 2 \large\frac{dy}{1+y^2}=-\int\frac{\tan ^{-1}(x)}{1+x^2}$$dx$
Consider $ \large\frac{dy}{1+y^2}$
This is of the form $ \int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1} \bigg(\large\frac{x}{a}\bigg)$
Hence $2 \int \large\frac{dy}{1+y^2}$$=2 \tan ^{-1}(y)$
Step 2:
Consider $\int \large\frac{-\tan ^{-1}(x)}{1+x^2}$$dx$
Let $\tan ^{-1}x=t$ on differentiating w.r.t x we get
$\large\frac{1}{1+x^2}$$dx=dt$
Now substituting this we get,
$\large\frac{-1}{2} $$\int t.dt$
On integrating we get
$-\bigg[\large\frac{t^2}{2}\bigg]$ substituting for t we get
$ =\large\frac{-(\tan ^{-1}x)^2}{2}$
Now applying this in required solution we get,
$ 2\tan ^{-1}y=\large\frac{-(\tan ^{-1}x)^2}{2}+c$
$ 2\tan ^{-1}y+\large\frac{(\tan ^{-1}x)^2}{2}=c$
Hence the required solution is
$ 2\tan ^{-1}y+\large\frac{1}{2}$$(\tan ^{-1}x)^2=c$