Question

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

Answer 1

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

$\int \frac{f^{\prime }\left(x\right)g\left(x\right)-g^{\prime }\left(x\right)f\left(x\right)dx}{\left[\frac{f\left(x\right)+1}{g\left(x\right)}\right]\left(g^2\right(x\left)\right)\sqrt{\frac{f\left(x\right)}{g\left(x\right)}-1}}Let:t^2=\frac{f\left(x\right)}{g\left(x\right)}-12tdt=\frac{f^{\prime }\left(x\right)g\left(x\right)-g^{\prime }\left(x\right)f\left(x\right)}{g^2\left(x\right)}\int \frac{2tdt}{(t^2+2)t}=\int \frac{2dt}{(t^2+2)}=\sqrt{2}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)=\sqrt{2}{\tan }^{-1}\left(\sqrt{\frac{f\left(x\right)-g\left(x\right)}{2g\left(x\right)}}\right)+C\sqrt{2}{\tan }^{-1}\left(\sqrt{\frac{f\left(x\right)-g\left(x\right)}{2g\left(x\right)}}\right)=\sqrt{m}{\tan }^{-1}\left(\sqrt{\frac{f\left(x\right)-g\left(x\right)}{ng\left(x\right)}}\right)m=2n=2m^2+n^2=2^2+2^2=8$

Hence the correct answer is $8$.