Answer
\begin{aligned} \int \tanh ^{-1}4 x \ dx =x\tanh ^{-1}4 x+\frac{1}{8} \ln|1-16x^2| +C\\
\end{aligned}
Work Step by Step
Given $$\int \tanh ^{-1}4 x \ dx $$ Use integration by parts: $$ u=\tanh ^{-1} 4x \Rightarrow du= \frac{4}{1-16x^2}dx $$ $$ dv= dx \Rightarrow v=x $$ So, we get \begin{aligned} I&=\int \tanh ^{-1}4 x \ dx\\
&=uv- \int vdu\\ &=x\tanh ^{-1}4 x- \int \frac{4x}{1-16x^2}d x\\ &=x\tanh ^{-1}4 x+\frac{1}{8} \int \frac{-32x}{1-16x^2} d x\\
&=x\tanh ^{-1}4 x+\frac{1}{8} \ln|1-16x^2| +C\\
\end{aligned}
