Question

Let $\int \frac{f^{\prime }\left(x\right)g\left(x\right)-g^{\prime }\left(x\right)f\left(x\right)}{\left(f\right(x)+g(x\left)\right)\sqrt{f\left(x\right)g\left(x\right)-g^2\left(x\right)}}dx=\sqrt{m}ta{n}^{-1}\left(\sqrt{\frac{f\left(x\right)-g\left(x\right)}{ng\left(x\right)}}\right)+C$,
where $m,n\in N$ and 'C' is constant of integration (g(x) > 0). Find the value of $(m^2+n^2)$.

Answer 1

$\int \frac{f^{{}^{\prime }}\left(x\right)g\left(x\right)-g^{{}^{\prime }}\left(x\right)f\left(x\right)}{\left(f\right(x)+g(x\left)\right)\sqrt{f\left(x\right)g\left(x\right)-g^2\left(x\right)}}dx=\int \frac{1}{(\frac{f\left(x\right)}{g\left(x\right)}+1)\sqrt{\left(\frac{f\left(x\right)}{g\left(x\right)}\right)-1}}\frac{f^{{}^{\prime }}\left(x\right)g\left(x\right)-g^{{}^{\prime }}\left(x\right)f\left(x\right)}{g^2\left(x\right)}dx$

Let $\frac{f\left(x\right)}{g\left(x\right)}-1=y^2$

$\Rightarrow \frac{f^{{}^{\prime }}\left(x\right)g\left(x\right)-g^{{}^{\prime }}\left(x\right)f\left(x\right)}{g^2\left(x\right)}dx=2ydy$

So, the integral becomes,

$\int \frac{2ydy}{y(y^2+2)}=2\int \frac{dy}{y^2+2}=2\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{y}{\sqrt{2}}\right)$

On Substituting for $y$,

$\int \frac{f^{{}^{\prime }}\left(x\right)g\left(x\right)-g^{{}^{\prime }}\left(x\right)f\left(x\right)}{\left(f\right(x)+g(x\left)\right)\sqrt{f\left(x\right)g\left(x\right)-g^2\left(x\right)}}dx=\sqrt{2}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\left(\sqrt{\frac{f\left(x\right)}{g\left(x\right)}-1}\right)\right)+C$

So, $m=2$ and $n=2$

ie, $m^2+n^2=4+4=8$