Question

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

• A. $1$
• B. $6$
• C. $4$
• D. $8$

Answer 1

The correct option is D $8$

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

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

Let $\frac{f\left(x\right)}{g\left(x\right)}-1=t^2$
$\Rightarrow \frac{f^{\prime }\left(x\right)g\left(x\right)-g^{\prime }\left(x\right)f\left(x\right)}{\left(g\right(x){)}^2}dx=2tdt\therefore I=\int \frac{2tdt}{(t^2+2)\cdot t}=\frac{2}{\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)+C=\sqrt{2}{\tan }^{-1}\sqrt{\frac{f\left(x\right)}{2g\left(x\right)}-\frac{1}{2}}+C=\sqrt{2}{\tan }^{-1}\sqrt{\frac{f\left(x\right)-g\left(x\right)}{2g\left(x\right)}}+C$

Hence $m=2$ and $n=2$
$\therefore m^2+n^2=8$