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Question Number 202866 by mnjuly1970 last updated on 04/Jan/24

$$$$$$calculate...$$$$\mathit{\varphi }={\int }_0^1\frac{{tanh}^{-1}\left(x\right)}{{(1+x)}^2}dx=?$$$$$$

Answered by Mathspace last updated on 06/Jan/24

$$x=th\left(t\right)=\frac{sht}{cht}=\frac{e^t-e^{-t}}{e^t+e^{-t}}$$$$\frac{dx}{dt}=\frac{{ch}^{2}t-{sh}^{2}t}{{ch}^{2}t}=\frac{1}{{ch}^{2}t}$$$$\mathrm{\Phi }={\int }_0^{argth1}\frac{t}{{(1+\frac{e^t-e^{-t}}{e^t+e^{-t}})}^2}\frac{dt}{{\left(\frac{e^t+e^{-t}}{2}\right)}^2}$$$$=4{\int }_0^{argth\left(1\right)}\frac{t}{4e^{2t}}dt$$$$={\int }_0^{argth\left(1\right)}t{e}^{-2t}dt$$$$={[-\frac{1}{2}t{e}^{-2t}]}_0^{argth\left(1\right)}+\frac{1}{2}{\int }_0^{argth1}{e}^{-2t}dt$$$$wehavex=\frac{e^t-e^{-t}}{e^t+e^{-t}}$$$$=\frac{e^{2t}-1}{e^{2t}+1}\Rightarrow {xe}^{2t}+x=e^{2t}-1\Rightarrow$$$$(x-1)e^{2t}=-x-1\Rightarrow$$$$e^{2t}=\frac{1+x}{1-x}\Rightarrow t=\frac{1}{2}ln\left(\frac{1+x}{1-x}\right)$$$$x\to 1^{-}\Rightarrow t\to +\mathrm{\infty }\Rightarrow$$$$\mathrm{\Phi }={[-\frac{1}{2}{te}^{-2t}]}_0^{\mathrm{\infty }}-\frac{1}{4}{\left[e^{-2t}\right]}_0^{\mathrm{\infty }}$$$$=\frac{1}{4}$$

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