∫_0 ^1 ((ln(x^2 +1))/(x+1))dx


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Question Number 96925 by M±th+et+s last updated on 05/Jun/20

$\int_{0}^{1} \frac{l n (x^{2} + 1)}{x + 1} d x$
	Answered by mathmax by abdo last updated on 06/Jun/20
	$let f(α) =∫_0 ^1 ((ln(x^2 +α))/(x+1))dx we have f(1) =∫_0 ^1 ((ln(x^2 +1))/(x+1))dx (α>0) f^′ (α) = ∫_0 ^1 (dx/((x^2 +α)(x+1))) let decompose F(x) =(1/((x+1)(x^2 +α))) F(x) =(a/(x+1)) +((bx+c)/(x^2 +α)) we have a =(1/(1+α)) lim_(x→+∞) xF(x) =0 =a+b ⇒b =−(1/(1+α)) F(0) =(1/α) =a+(c/α) ⇒1 =αa +c ⇒c =1−αa =1−(α/(1+α)) =(1/(1+α)) ⇒ F(x) =(1/((1+α)(x+1))) +((−(1/(1+α))x +(1/(1+α)))/(x^2 +α)) =(1/(1+α)){ (1/(x+1))−(x/(x^2 +α)) +(1/(x^2 +α))} ⇒ ∫ F(x) =(1/(1+α)){ln∣x+1∣−(1/2)ln(x^2 +α) +∫ (dx/(x^2 +α))} ∫ (dx/(x^2 +α)) =_(x =(√α)u) ∫ (((√α)du)/(α(1+u^2 ))) =(1/(√α)) arctan((x/(√α))) ⇒ ∫_0 ^1 F(x)dx =(1/(1+α)){ [ln∣((x+1)/(√(x^2 +α)))∣]_0 ^1 +[(1/(√α)) arctan((x/(√α)))]_0 ^1 } =(1/(1+α)){ ln((2/(√(1+α))))−ln((1/(√α))) +(1/(√α)) arctan((1/(√α)))} ⇒ f(α) =∫ ((ln((2/((√(1+α)) ))))/(1+α)) dα +(1/2)∫ ((lnα)/(1+α))dα +∫ ((arctan((1/(√α))))/((1+α)(√α))) dα +c ...be continued....$
	$let f (α) = \int_{0}^{1} \frac{\ln (x^{2} + α)}{x + 1} dx we have f (1) = \int_{0}^{1} \frac{\ln (x^{2} + 1)}{x + 1} dx (α > 0)$ $f^{^{'}} (α) = \int_{0}^{1} \frac{dx}{(x^{2} + α) (x + 1)} let decompose F (x) = \frac{1}{(x + 1) (x^{2} + α)}$ $F (x) = \frac{a}{x + 1} + \frac{bx + c}{x^{2} + α}$ $we have a = \frac{1}{1 + α}$ $\lim_{x \to + \infty} xF (x) = 0 = a + b \Rightarrow b = - \frac{1}{1 + α}$ $F (0) = \frac{1}{α} = a + \frac{c}{α} \Rightarrow 1 = α a + c \Rightarrow c = 1 - α a = 1 - \frac{α}{1 + α} = \frac{1}{1 + α} \Rightarrow$ $F (x) = \frac{1}{(1 + α) (x + 1)} + \frac{- \frac{1}{1 + α} x + \frac{1}{1 + α}}{x^{2} + α} = \frac{1}{1 + α} {\frac{1}{x + 1} - \frac{x}{x^{2} + α} + \frac{1}{x^{2} + α}} \Rightarrow$ $\int F (x) = \frac{1}{1 + α} {\ln ∣ x + 1 ∣ - \frac{1}{2} \ln (x^{2} + α) + \int \frac{dx}{x^{2} + α}}$ $\int \frac{dx}{x^{2} + α} =_{x = \sqrt{α} u} \int \frac{\sqrt{α} du}{α (1 + u^{2})} = \frac{1}{\sqrt{α}} \arctan (\frac{x}{\sqrt{α}}) \Rightarrow$ $\int_{0}^{1} F (x) dx = \frac{1}{1 + α} {{[\ln ∣ \frac{x + 1}{\sqrt{x^{2} + α}} ∣]}_{0}^{1} + {[\frac{1}{\sqrt{α}} \arctan (\frac{x}{\sqrt{α}})]}_{0}^{1}}$ $= \frac{1}{1 + α} {\ln (\frac{2}{\sqrt{1 + α}}) - \ln (\frac{1}{\sqrt{α}}) + \frac{1}{\sqrt{α}} \arctan (\frac{1}{\sqrt{α}})} \Rightarrow$ $f (α) = \int \frac{\ln (\frac{2}{\sqrt{1 + α}})}{1 + α} d α + \frac{1}{2} \int \frac{\ln α}{1 + α} d α + \int \frac{\arctan (\frac{1}{\sqrt{α}})}{(1 + α) \sqrt{α}} d α + c$ $. . . be continued . . . .$
	Answered by mathmax by abdo last updated on 07/Jun/20

	$let take a try with series we have$ $I = \int_{0}^{1} \ln (1 + x^{2}) \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} dx = \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{n} \ln (1 + x^{2}) dx$ $A_{n} = \int_{0}^{1} x^{n} \ln (1 + x^{2}) dx we have \ln^{^{'}} (1 + u) = \frac{1}{1 + u} = \sum_{p = 0}^{\infty} {(- 1)}^{p} u^{p} \Rightarrow$ $\ln (1 + u) = \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p} u^{p + 1}}{p + 1} + c (c = 0)$ $= \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1} u^{p}}{p} \Rightarrow \ln (1 + x^{2}) = \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p} x^{2 p} \Rightarrow$ $A_{n} = \int_{0}^{1} x^{n} \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p} x^{2 p} dx = \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p} \int_{0}^{1} x^{n + 2 p} dx$ $= \sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p (n + 2 p + 1)} \Rightarrow I = \sum_{n = 0}^{\infty} {(- 1)}^{n} (\sum_{p = 1}^{\infty} \frac{{(- 1)}^{p - 1}}{p (n + 2 p + 1)})$ $I = \sum_{n ⩾ 0} \sum_{p ⩾ 1} \frac{{(- 1)}^{n + p - 1}}{p (n + 2 p + 1)}$
	Commented by M±th+et+s last updated on 07/Jun/20

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