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


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Question Number 112203 by mathdave last updated on 06/Sep/20

$p r o p o r s e d b y m . n j u l y 1970$ $\int_{0}^{1} \frac{\ln (\sqrt[3]{x + 1})}{(x + 1) (x + 2)} d x$ $s o l u t i o n$ $l e t I = \frac{1}{3} \int_{0}^{1} \frac{\ln (x + 1)}{(x + 1) (x + 2)} d x$ $I = \frac{1}{3} \int_{0}^{1} \frac{\ln (1 + x)}{x + 1} d x - \int_{0}^{1} \frac{\ln (x + 1)}{x + 2} d x = A - B$ $l e t A = \frac{1}{3} \int_{0}^{1} \frac{\ln (x + 1)}{x + 1} d x (u s i n g I B P)$ $A = \frac{1}{3} {[\ln^{2} (x + 1)]}_{0}^{1} - \frac{1}{3} \int_{0}^{1} \frac{\ln (x + 1)}{x + 1} d x$ $A (1 + \frac{1}{3}) = \frac{1}{3} \ln^{2} (2)$ $A = \int_{0}^{1} \frac{\ln (x + 1)}{x + 1} d x = \frac{1}{3} \ln^{2} (2) . . . . . . (1)$ $t h e n B = \frac{1}{3} \int_{0}^{1} \frac{\ln (x + 1)}{x + 2} d x$ $B = \frac{1}{3} {[\ln (x + 2) \ln (x + 1)]}_{0}^{1} - \frac{1}{3} \int_{0}^{1} \frac{\ln (x + 2)}{x + 1} d x$ $l e t x + 1 = t$ $B = \frac{1}{3} \ln 3 \ln 2 - \frac{1}{3} \int_{1}^{2} \frac{\ln (1 + t)}{t} d t$ $l e t t = - x$ $B = \frac{1}{3} \ln 3 \ln 2 - \frac{1}{3} \int_{- 1}^{- 2} \frac{\ln (1 - x)}{x} d x$ $b u t \int \frac{\ln (1 - x)}{x} d x = - {L i}_{2} (x)$ $B = \frac{1}{3} \ln 3 \ln 2 + \frac{1}{3} {[{L i}_{2} (x)]}_{- 1}^{- 2}$ $B = \frac{1}{3} \ln 3 \ln 2 + \frac{1}{3} {L i}_{2} (- 2) - \frac{1}{3} {L i}_{2} (- 1)$ $B = \frac{1}{3} \ln 3 \ln 2 + \frac{1}{3} {L i}_{2} (- 2) + \frac{π^{2}}{36}$ $b u t I = A - B$ $\int_{0}^{1} \frac{\ln (\sqrt[3]{1 + x})}{(x + 1) (x + 2)} d x = \frac{1}{4} \ln^{2} (2) - \frac{1}{3} \ln 3 \ln 2 - \frac{1}{3} {L i}_{2} (- 2) - \frac{π^{2}}{36}$ $m a t h d a v e (06 / 09 / 2020)$
Commented by mnjuly1970 last updated on 06/Sep/20

$v e r y n i c e, t h a n k y o u s i r$ $m a t h d a v e . g r a t e f u l f o r y o u r$ $e f f o r t a n d f a v o r . . . .$
Commented by Tawa11 last updated on 06/Sep/21

$great sir$
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