Answer
\begin{aligned} \int \sinh ^{-1} x \ dx=x\sinh ^{-1} x- \sqrt{x^2+1} +C\\
\end{aligned}
Work Step by Step
Given $$\int \sinh ^{-1} x \ dx $$ Use integration by parts:$$ u=\sinh ^{-1} x \Rightarrow du= \frac{1}{\sqrt{x^2+1}}dx $$ $$ dv= dx \Rightarrow v=x $$ So, we get \begin{aligned} I&=\int \sinh ^{-1} x \ dx\\
&=uv- \int vdu\\ &=x\sinh ^{-1} x- \int \frac{x}{\sqrt{x^2+1}}dx\\
& =x\sinh ^{-1} x-\frac{1}{2} \int \frac{2x}{\sqrt{x^2+1}}dx\\
& =x\sinh ^{-1} x-\frac{1}{2} \int 2x (x^2+1)^{-\frac{1}{2}}dx\\
& =x\sinh ^{-1} x- (x^2+1)^{\frac{1}{2}} +C\\
& =x\sinh ^{-1} x- \sqrt{x^2+1} +C\\
\end{aligned}
