If $$\displaystyle \frac {dy}{dx} = - \frac {x(1+y^{2})^{2}}{\ln y (1-y^{2})}$$
then
$$\displaystyle \lim_{y \to \infty} x(y) = \sqrt{\frac {A \pi}{B} + C}$$
where $A$and $B$ are coprime positive integers, find the value of $A + B + C$.
ATTEMPT
Used separation of variables to integrate for xx: $$\frac {dy}{dx} = - \frac {x\,(1+y^{2})^{2}}{\ln(y) \,(1 - y^{2})}$$ $$\therefore \int x\,\mathrm{d}x = - \int \ln(y)\,\frac {1-y^{2}}{(1+y^{2})^{2}} \,\mathrm{d}y$$
$$\frac {1}{2}x^{2} = - \int \ln(y)\,\frac {1-y^{2}}{(1+y^{2})^{2}} \,\mathrm{d}y$$
Now used integration by parts to evaluate the right-hand-side where $f'(y)$ = $\frac{1-y^{2}}{(1+y^{2})^{2}}$
$g(y) = \ln(y)g(y)=ln(y$) and hence $g'(y) = \frac {1}{y}g$
To find $f(y)f(y)$ use partial fractions where $z=y^{2}$ : $$\frac {1-z}{(1+z)^{2}} = \frac {A}{(1+z)^{2}} + \frac {B}{1+z}$$
$$1-z = {A} + B\,(1+z)$$
When $z = -1$ $A = 2$ hence $B = -1$for all $z$.
Rewriting the expression for $f(y)f(y)$, now used the substitution $u$ = $arctany$ and $\mathrm{d}y = \sec^{2}u\mathrm{d}$ for the integral on the left side.
$$f(y) = \int \frac {2}{(1+y^{2})^{2}}\mathrm{d}y-\int \frac {1}{1+y^{2}}\,\mathrm{d}y$$
$$f(y) = \int \frac {2}{(1+\tan^{2}u)^{2}\,\sec^{2}(u)}\mathrm{d}u - \arctan y$$
$$f(y) = \int 2\cos^{2}(u)\mathrm{d}u - \arctan y$$
Used the identity $\cos 2u = 2\cos^{2}$ to simplify: $$f(y) = \int \cos 2u +1 \,\mathrm{d}u - \arctan y$$ = $$\frac {1}{2} \sin 2u + u - \arctan y$$1
= $$\sin u \cos u + u - \arctan y$$
=$$ \sin (\arctan y) \cos (\arctan y) + \arctan y - \arctan y$$
Used the identities $\sin (\arctan y) = \frac {y}{\sqrt{1+y^{2}}}sin(arctany)$=
and $\cos (\arctan y) = \frac {1}{\sqrt{1+y^{2}}}cos(arctany)$
to obtain $$f(y) = \frac {y}{1+y^{2}}$$
then
$$\frac {1}{2}x^{2} = - \int \ln(y)\,\frac {1-y^{2}}{(1+y^{2})^{2}} \,\mathrm{d}y$$2
$$\frac {1}{2}x^{2} = - \frac {y}{1+y^{2}}\,\ln y + \int \frac {1}{y}.\frac {y}{1+y^{2}} \,\mathrm{d}y$$
$$\frac {1}{2}x^{2} = - \frac {y}{1+y^{2}}\,\ln y + \arctan y + cy$$
Substituted the point (1, 1) to obtain the constant of integration: $$\frac {1}{2} = \frac {\pi}{4} + c$$
$$c = \frac {1}{2} - \frac {\pi}{4}$$
$$\frac {1}{2}x^{2} = -\frac {y}{1+y^{2}}\,\ln y + \arctan y + \frac {1}{2} - \frac {\pi}{4}$$π
$$x = \pm \sqrt{-\frac {2y}{1+y^{2}}\,\ln y + 2\arctan y + 1 - \frac {\pi}{2}}$$
To satisfy the condition y(1)=1y(1)=1, the negative result is ignored. $$x = \sqrt{-\frac {2y}{1+y^{2}}\,\ln y + 2\arctan y + 1 - \frac {\pi}{2}}$$
$$\lim_{y \to \infty} x(y) = \lim_{y \to \infty} \sqrt{-\frac {2y}{1+y^{2}}\,\ln y + 2\arctan y + 1 - \frac {\pi}{2}}$$
As $$y \to \infty→ -\frac {2y}{1+y^{2}}$$ $$lny→0$$ and $$2\arctan y \to \pi2arctany→π$$
Hence the limit becomes: $$\sqrt{0 + \pi + 1 - \frac {\pi}{2}} = \sqrt{\frac {\pi}{2} + 1}$$
$$\therefore A + B + C = \boxed{4}$$
Is my solution correct?
Please suggest any other way because it is too lengthy