Question Number 176594 by mnjuly1970 last updated on 22/Sep/22
![𝛗=∫_0 ^( 1) (( ( tanh^( −1) (x))^2 )/((1+x )^( 2) )) dx = ? ≺ solution ≻ note : tanh^( −1) (x)=− (1/2) ln(((1−x)/(1+x))) 𝛗= (1/4)∫_0 ^( 1) (( ln^( 2) (((1−x)/(1+x)) ))/((1+x )^( 2) )) dx =^(((1−x)/(1+x)) = t) (1/8)∫_0 ^( 1) ln^( 2) (t )dt =(1/8_ ) { [t.ln^( 2) (t)]_0 ^( 1) −2∫_0 ^( 1) ln(t)dt} =− (1/4) ∫_0 ^( 1) ln(t)dt= (1/4) ◂ m.n ▶](Q176594.png)
$$ \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\left(\:{tanh}^{\:−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}+{x}\:\right)^{\:\mathrm{2}} }\:{dx}\:=\:?\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\prec\:\:\:{solution}\:\:\succ \\ $$$$\:\:\:\:\:{note}\::\:\:{tanh}^{\:−\mathrm{1}} \left({x}\right)=−\:\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right) \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:\right)}{\left(\mathrm{1}+{x}\:\right)^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\overset{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:=\:{t}} {=}\:\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}^{\:\mathrm{2}} \left({t}\:\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8}_{\:} }\:\left\{\:\left[{t}.{ln}^{\:\mathrm{2}} \left({t}\right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right){dt}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right){dt}=\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\blacktriangleleft\:{m}.{n}\:\blacktriangleright\: \\ $$
Commented by Tawa11 last updated on 23/Sep/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$