Question

$\int {\sinh }^{-1}\left(\frac{x}{4}\right)dx$ is equal to

• A. $x{\sinh }^{-1}\left(\frac{x}{4}\right)-\sqrt{x^2+16}+c$
• B. $x{\sinh }^{-1}\left(\frac{x}{4}\right)+\sqrt{x^2+16}+c$
• C. $x{\sinh }^{-1}\left(\frac{x}{4}\right)-\frac{1}{2}\sqrt{x^2+16}+c$
• D. $x{\sinh }^{-1}\left(\frac{x}{2}\right)-x\sqrt{x^2+16}+c$

Answer 1

The correct option is A $x{\sinh }^{-1}\left(\frac{x}{4}\right)-\sqrt{x^2+16}+c$

Let $I\int {\sinh }^{-1}\left(\frac{x}{4}\right)dx$

Put $\frac{x}{4}=\sinh \theta$

$⟹dx=4\cosh \theta d\theta$

$I=\int \theta 4\cosh \theta d\theta =4\int \theta \cosh \theta d\theta$

$=4\theta \sinh \theta -\int 1\cdot \sinh \theta d\theta$ ....... integration by parts

$=4\theta \sinh \theta -4\cosh \theta +c$

$=x{\sinh }^{-1}\left(\frac{x}{4}\right)-\sqrt{x^2+16}+c$