Question Number 176334 by mnjuly1970 last updated on 16/Sep/22
$$ \\ $$$$\:\:\:\:\:\Psi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{ln}\left(\:\mathrm{1}+\:{x}\:−\:{x}^{\:\mathrm{2}} \right)}{{x}}\mathrm{d}{x}\:=\:? \\ $$$$ \\ $$
Answered by Peace last updated on 20/Sep/22
![1+x−x^2 =((1−x^3 )/(1+x)) Ψ=∫_0 ^1 ((ln(1−x^3 ))/x)dx_ −∫_0 ^1 ((ln(1+x))/x)dx_(x=−t) Ψ=A−∫_0 ^(−1) ((ln(1−t))/t)dt=A+Li_2 (−1) A=∫_0 ^1 ((ln(1−x^3 ))/x)dx=(1/3)∫_0 ^1 ((ln(1−x^3 ))/x^3 )dx^3 =−(1/3)[Li_2 (x^3 )]_0 ^1 =((Li_2 (1))/3) Ψ=((Li_2 (1))/3)+Li_2 (−1)](https://www.tinkutara.com/question/Q176530.png)
$$\mathrm{1}+{x}−{x}^{\mathrm{2}} =\frac{\mathrm{1}−{x}^{\mathrm{3}} }{\mathrm{1}+{x}} \\ $$$$\Psi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{3}} \right)}{{x}}{dx}_{} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}_{{x}=−{t}} \\ $$$$\Psi={A}−\int_{\mathrm{0}} ^{−\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}={A}+{Li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{3}} \right)}{{x}}{dx}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{3}} }{dx}^{\mathrm{3}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\left[{Li}_{\mathrm{2}} \left({x}^{\mathrm{3}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{{Li}_{\mathrm{2}} \left(\mathrm{1}\right)}{\mathrm{3}} \\ $$$$\Psi=\frac{{Li}_{\mathrm{2}} \left(\mathrm{1}\right)}{\mathrm{3}}+{Li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$