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Question Number 139655 by mohammad17 last updated on 30/Apr/21

	Answered by qaz last updated on 30/Apr/21
	$∫_0 ^1 xtan^(−1) (x)ln(x^2 +1)dx =(1/2)∫_0 ^1 tan^(−1) (x)ln(x^2 +1)d(x^2 +1) =(1/2)∫_0 ^1 tan^(−1) (x)d[(x^2 +1)ln(x^2 +1)−(x^2 +1)] =(1/2){(π/4)(2ln2−2)−∫_0 ^1 (([(x^2 +1)ln(x^2 +1)−(x^2 +1)])/(x^2 +1))dx} =(1/2){(π/2)ln(2/e)−∫_0 ^1 [ln(x^2 +1)−1]dx} =(π/4)ln(2/e)+(1/2)−(1/2){ln2−∫_0 ^1 ((2x^2 )/(x^2 +1))dc} =(π/4)ln(2/e)+(1/2)−(1/2)ln2+1−(π/4) =(π/4)ln(2/e^2 )+(3/2)−(1/2)ln2$
	$\int_{0}^{1} x \tan^{- 1} (x) l n (x^{2} + 1) d x$ $= \frac{1}{2} \int_{0}^{1} \tan^{- 1} (x) l n (x^{2} + 1) d (x^{2} + 1)$ $= \frac{1}{2} \int_{0}^{1} \tan^{- 1} (x) d [(x^{2} + 1) l n (x^{2} + 1) - (x^{2} + 1)]$ $= \frac{1}{2} {\frac{π}{4} (2 l n 2 - 2) - \int_{0}^{1} \frac{[(x^{2} + 1) l n (x^{2} + 1) - (x^{2} + 1)]}{x^{2} + 1} d x}$ $= \frac{1}{2} {\frac{π}{2} l n \frac{2}{e} - \int_{0}^{1} [l n (x^{2} + 1) - 1] d x}$ $= \frac{π}{4} l n \frac{2}{e} + \frac{1}{2} - \frac{1}{2} {l n 2 - \int_{0}^{1} \frac{2 x^{2}}{x^{2} + 1} d c}$ $= \frac{π}{4} l n \frac{2}{e} + \frac{1}{2} - \frac{1}{2} l n 2 + 1 - \frac{π}{4}$ $= \frac{π}{4} l n \frac{2}{e^{2}} + \frac{3}{2} - \frac{1}{2} l n 2$
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