Question

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

• A. ${\sin }^{-1}\sqrt{\frac{f\left(x\right)}{\varphi \left(x\right)}}+c$
• B. ${\cos }^{-1}\sqrt{\left(f\right(x){)}^2-(\varphi \left(x\right){)}^2}+c$
• C. $\sqrt{2}{\tan }^{-1}\sqrt{\frac{f\left(x\right)\varphi \left(x\right)-1}{2}}+c$
• D. $\sqrt{2}{\tan }^{-1}\sqrt{\frac{f\left(x\right)\varphi \left(x\right)+1}{2}}+c$

Answer 1

The correct option is C $\sqrt{2}{\tan }^{-1}\sqrt{\frac{f\left(x\right)\varphi \left(x\right)-1}{2}}+c$

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

Let $f\left(x\right)\varphi \left(x\right)-1=t^2$
$\Rightarrow f^{\prime }\left(x\right)\varphi \left(x\right)+f\left(x\right){\varphi }^{\prime }\left(x\right)=2tdt$

$\therefore I=\int \frac{2tdt}{(t^2+2)\left(t\right)}\Rightarrow I=2\int \frac{dt}{t^2+2}\Rightarrow I=\frac{2}{\sqrt{2}}{\tan }^{-1}\frac{t}{\sqrt{2}}+c\Rightarrow I=\sqrt{2}{\tan }^{-1}\left(\sqrt{\frac{f\left(x\right)\varphi \left(x\right)-1}{2}}\right)+c$