## Calculus (3rd Edition)

$$\frac{1}{2} x^{2} \tanh ^{-1} x+\frac{1}{2}x+\frac{1}{4}\ln|1-x|-\frac{1}{4}\ln|1+x|+C$$
Given $$\int x \tanh ^{-1} x d x$$Let \begin{align*} u&= \tanh ^{-1} x \ \ \ \ \ \ \ \ \ dv=x dx\\ du&= \frac{1}{1-x^2}dx \ \ \ \ \ \ \ \ \ dv=\frac{1}{2}x^{2} \end{align*} Then \begin{align*} \int x \tanh ^{-1} x d x &= \frac{1}{2} x^{2} \tanh ^{-1} x-\frac{1}{2} \int \frac{x^{2}}{1-x^{2}} d x\\ &=\frac{1}{2} x^{2} \tanh ^{-1} x-\frac{1}{2} \int\left[-1+\frac{1}{1-x^{2}}\right] d x\\ &=\frac{1}{2} x^{2} \tanh ^{-1} x+\frac{1}{2} \int 1 \, d x-\frac{1}{2} \int \frac{1}{1-x^{2}} d x \end{align*} Since \begin{aligned} \frac{1}{1-x^{2}} &=\frac{1}{(1-x)(1+x)} \\ &=\frac{A}{(1-x)}+\frac{B}{(1+x)} \\ 1 &=A(1+x)+B(1-x) \end{aligned} At $x=1$, $A=1/2$ and at $x= -1$, $B= 1/2$, so \begin{align*} \int \frac{1}{1-x^{2}} d x &=\frac{1}{2}\int \frac{dx}{(1-x)}+\frac{1}{2}\int \frac{dx}{(1+x)}\\ &=-\frac{1}{2}\ln|1-x|+\frac{1}{2}\ln|1+x|+C \end{align*} Hence $$\int x \tanh ^{-1} x d x = \frac{1}{2} x^{2} \tanh ^{-1} x+\frac{1}{2}x+\frac{1}{4}\ln|1-x|-\frac{1}{4}\ln|1+x|+C$$