Question

$\int \frac{1}{(1+x^2)\sqrt{p^2+q^2{\left({\tan }^{-1}x\right)}^2}}dx$ is equal to

• A. $\frac{1}{p}\log |p\left({\tan }^{-1}x\right)+\sqrt{p^2+{\left(q{\tan }^{-1}x\right)}^2}|+C$
• B. $\frac{1}{p}\log |\left(p{\cot }^{-1}x\right)+\sqrt{p^2+{\left(p{\cot }^{-1}x\right)}^2}|+C$
• C. $\frac{1}{q}\log |q\left({\cot }^{-1}x\right)+\sqrt{p^2+{\left(q{\cot }^{-1}x\right)}^2}|+C$
• D. $\frac{1}{q}\log |q\left({\tan }^{-1}x\right)+\sqrt{p^2+{\left(q{\tan }^{-1}x\right)}^2}|+C$

Answer 1

Answer: (D)

$\frac{1}{q}\log |q\left({\tan }^{-1}x\right)+\sqrt{p^2+{\left(q{\tan }^{-1}x\right)}^2}|+C$

$I=\int \frac{dx}{(x^2+1)\sqrt{p^2+q^2{\left({\tan }^{-1}x\right)}^2}}$
Put $\left(q{\tan }^{-1}x\right)=t⟹\frac{q}{1+x^2}dx=dt$
$\therefore I=\frac{1}{q}\int \frac{dt}{\sqrt{p^2+t^2}}$
$=\frac{1}{q}\log |t+\sqrt{p^2+t^2}|+C$
$=\frac{1}{q}\log |q\left({\tan }^{-1}x\right)+\sqrt{p^2+{\left(q{\tan }^{-1}x\right)}^2}|+C$