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prove-0-1-1-x-2-ln-3-1-x-x-dx-51-8-pi-4-15-m-n-

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Question Number 166082 by mnjuly1970 last updated on 12/Feb/22
prove Ω = ∫_0 ^( 1) (( (1−x )^( 2) .ln^( 3) (1−x ))/x) dx = ((51)/8) −(π^( 4) /(15)) ■ m.n
$$ \\ $$$$\:\:\:\:\:\:\:{prove} \\ $$$$\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\left(\mathrm{1}−{x}\:\right)^{\:\mathrm{2}} .{ln}^{\:\mathrm{3}} \left(\mathrm{1}−{x}\:\right)}{{x}}\:{dx}\:=\:\frac{\mathrm{51}}{\mathrm{8}}\:−\frac{\pi^{\:\mathrm{4}} }{\mathrm{15}}\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:{m}.{n} \\ $$$$\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by qaz last updated on 13/Feb/22
∫_0 ^1 (((l−x)^2 ln^3 (1−x))/x)dx =∫_0 ^1 ((x^2 ln^3 x)/(1−x))dx =−∫_0 ^1 (1+x)ln^3 xdx+∫_0 ^1 ((ln^3 x)/(1−x))dx =−(x+(1/2)x^2 )ln^3 x∣_0 ^1 +3∫_0 ^1 (1+(1/2)x)ln^2 xdx−Σ_(n=0) ^∞ (6/((n+1)^4 )) =3(x+(1/4)x^2 )ln^2 x∣_0 ^1 −6∫_0 ^1 (1+(1/4)x)lnxdx−(π^4 /(15)) =−6(x+(1/8)x^2 )lnx∣_0 ^1 +6∫_0 ^1 (1+(1/8)x)dx−(π^4 /(15)) =6(x+(1/(16))x^2 )∣_0 ^1 −(π^4 /(15)) =((51)/8)−(π^4 /(15))
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{l}−\mathrm{x}\right)^{\mathrm{2}} \mathrm{ln}^{\mathrm{3}} \left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{ln}^{\mathrm{3}} \mathrm{x}}{\mathrm{1}−\mathrm{x}}\mathrm{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{x}\right)\mathrm{ln}^{\mathrm{3}} \mathrm{xdx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{3}} \mathrm{x}}{\mathrm{1}−\mathrm{x}}\mathrm{dx} \\ $$$$=−\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)\mathrm{ln}^{\mathrm{3}} \mathrm{x}\mid_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)\mathrm{ln}^{\mathrm{2}} \mathrm{xdx}−\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{6}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\mathrm{3}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} \right)\mathrm{ln}^{\mathrm{2}} \mathrm{x}\mid_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{6}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}\right)\mathrm{lnxdx}−\frac{\pi^{\mathrm{4}} }{\mathrm{15}} \\ $$$$=−\mathrm{6}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}^{\mathrm{2}} \right)\mathrm{lnx}\mid_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{6}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}\right)\mathrm{dx}−\frac{\pi^{\mathrm{4}} }{\mathrm{15}} \\ $$$$=\mathrm{6}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{x}^{\mathrm{2}} \right)\mid_{\mathrm{0}} ^{\mathrm{1}} −\frac{\pi^{\mathrm{4}} }{\mathrm{15}} \\ $$$$=\frac{\mathrm{51}}{\mathrm{8}}−\frac{\pi^{\mathrm{4}} }{\mathrm{15}} \\ $$
Commented by mnjuly1970 last updated on 13/Feb/22
thx alot sir qaz
$${thx}\:{alot}\:{sir}\:{qaz} \\ $$

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