Can someone help check if I evaluated my integral right?

Question

The integral given to be evaluated is $\int 6x\cdot\mathrm{arctanh}(x)\,dx$.

I tried to evaluated and got the following answer:

$$3x^2\operatorname{arctanh}x + 3x -3\ln\,\bigl\lvert(1+x)/\sqrt{1-x^2}\bigr\rvert + c$$

I used some trig substitutions and integration by parts to find it. Did I get it right or no?

Thank you.

Answer 1

I get:

$3 \left(\left(x^2-1\right) \tanh ^{-1}(x)+x\right) + C$.

Answer 2

Consider the function $y = \tanh^{-1} x$; that is to say, $$x = \tanh y = \frac{\sinh y}{\cosh y} = \frac{e^y - e^{-y}}{e^y + e^{-y}}.$$ Hence $$0 = x(e^y + e^{-y}) - e^y + e^{-y} = (x-1)e^y + (x+1)e^{-y},$$ or equivalently, $$e^{2y} = \frac{1+x}{1-x}.$$ Therefore, $$y = \tanh^{-1} x = \frac{1}{2} \log \frac{1+x}{1-x}, \quad |x| < 1.$$ Now observe that $$\begin{align*} \log \left| \frac{1+x}{\sqrt{1-x^2}} \right| &= \log \left| \frac{1+x}{\sqrt{1-x} \sqrt{1+x}} \right| = \log \left| \frac{\sqrt{1+x}}{\sqrt{1-x}} \right| = \log \left| \sqrt{\frac{1+x}{1-x}} \right| = \frac{1}{2} \log \left| \frac{1+x}{1-x} \right| \\ &= \tanh^{-1} x. \end{align*}$$ This immediately leads to the desired simplification.