Question Number 112203 by mathdave last updated on 06/Sep/20
![proporsed by m.njuly 1970 ∫_0 ^1 ((ln(((x+1))^(1/3) ))/((x+1)(x+2)))dx solution let I=(1/3)∫_0 ^1 ((ln(x+1))/((x+1)(x+2)))dx I=(1/3)∫_0 ^1 ((ln(1+x))/(x+1))dx−∫_0 ^1 ((ln(x+1))/(x+2))dx=A−B let A=(1/3)∫_0 ^1 ((ln(x+1))/(x+1))dx ( using IBP) A=(1/3)[ln^2 (x+1)]_0 ^1 −(1/3)∫_0 ^1 ((ln(x+1))/(x+1))dx A(1+(1/3))=(1/3)ln^2 (2) A=∫_0 ^1 ((ln(x+1))/(x+1))dx=(1/3)ln^2 (2)......(1) then B=(1/3)∫_0 ^1 ((ln(x+1))/(x+2))dx B=(1/3)[ln(x+2)ln(x+1)]_0 ^1 −(1/3)∫_0 ^1 ((ln(x+2))/(x+1))dx let x+1=t B=(1/3)ln3ln 2−(1/3)∫_1 ^2 ((ln(1+t))/t)dt let t=−x B=(1/3)ln3ln2−(1/3)∫_(−1) ^(−2) ((ln(1−x))/x)dx but ∫((ln(1−x))/x)dx=−Li_2 (x) B=(1/3)ln3ln2+(1/3)[Li_2 (x)]_(−1) ^(−2) B=(1/3)ln3ln2+(1/3)Li_2 (−2)−(1/3)Li_2 (−1) B=(1/3)ln3ln2+(1/3)Li_2 (−2)+(π^2 /(36)) but I=A−B ∫_0 ^1 ((ln(((1+x))^(1/3) ))/((x+1)(x+2)))dx=(1/4)ln^2 (2)−(1/3)ln3ln2−(1/3)Li_2 (−2)−(π^2 /(36)) mathdave(06/09/2020)](Q112203.png)
$${proporsed}\:{by}\:{m}.{njuly}\:\mathrm{1970} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx} \\ $$$${solution} \\ $$$${let}\:{I}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}+\mathrm{1}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{2}}{dx}={A}−{B} \\ $$$${let}\:{A}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}}{dx}\:\left(\:{using}\:{IBP}\right) \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{ln}^{\mathrm{2}} \left({x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}}{dx} \\ $$$${A}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)......\left(\mathrm{1}\right) \\ $$$${then}\:{B}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{2}}{dx} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{ln}\left({x}+\mathrm{2}\right)\mathrm{ln}\left({x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left({x}+\mathrm{2}\right)}{{x}+\mathrm{1}}{dx} \\ $$$${let}\:{x}+\mathrm{1}={t} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln}\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{ln}\left(\mathrm{1}+{t}\right)}{{t}}{dt} \\ $$$${let}\:{t}=−{x} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln2}−\frac{\mathrm{1}}{\mathrm{3}}\int_{−\mathrm{1}} ^{−\mathrm{2}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx} \\ $$$${but}\:\int\frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}=−{Li}_{\mathrm{2}} \left({x}\right) \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln2}+\frac{\mathrm{1}}{\mathrm{3}}\left[{Li}_{\mathrm{2}} \left({x}\right)\right]_{−\mathrm{1}} ^{−\mathrm{2}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln2}+\frac{\mathrm{1}}{\mathrm{3}}{Li}_{\mathrm{2}} \left(−\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{3}}{Li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln2}+\frac{\mathrm{1}}{\mathrm{3}}{Li}_{\mathrm{2}} \left(−\mathrm{2}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$${but}\:{I}={A}−{B} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+{x}}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln3ln2}−\frac{\mathrm{1}}{\mathrm{3}}{Li}_{\mathrm{2}} \left(−\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$${mathdave}\left(\mathrm{06}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 06/Sep/20
$${very}\:{nice},\:{thank}\:{you}\:{sir}\:\: \\ $$$${mathdave}\:.{grateful}\:{for}\:{your} \\ $$$${effort}\:\:{and}\:{favor}.... \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$