∫_( 0) ^( 1) ((x^3 − 1)/((1 + x^2 ) ln x)) dx


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Question Number 52944 by Tawa1 last updated on 15/Jan/19

$\int_{0}^{1} \frac{x^{3} - 1}{(1 + x^{2}) \ln x} dx$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19
	$I(a)=∫_0 ^1 ((x^a −1)/(1+x^2 ))×(dx/(lnx)) (dI_ /da)=∫_0 ^1 (∂/∂a)(((x^a −1)/(1+x^2 )))×(dx/(lnx)) =∫_0 ^1 ((x^a lnx)/(1+x^2 ))×(dx/(lnx)) =∫_0 ^1 (x^a /(1+x^2 ))dx =∫_0 ^1 x^a (1−x^2 +x^4 −x^6 +...)dx =∫_0 ^1 x^a −x^(2+a) +x^(4+a) −x^(6+a) +...dx =∣(x^(a+1) /(a+1))−(x^(a+3) /(a+3))−(x^(a+5) /(a+5))−(x^(a+7) /(a+7))....∣_0 ^1 =(1/(a+1))−(1/(a+3))+(1/(a+5))−(1/(a+7))+... now (dI/da)=(1/(a+1))−(1/(a+3))+(1/(a+5))−(1/(a+7))+... ∫dI=∫((1/(a+1))−(1/(a+3))+(1/(a+5))−(1/(a+7))+...)da I(a)=[ln(a+1)−ln(a+3)+ln(a+5)−...]+c I(a)=∫_0 ^1 ((x^a −1)/((1+x^2 )lnx))dx when a=0 I(a)=0 so 0=[ln1−ln3+ln5−ln7...]+c c=−ln1+ln3−ln5+ln7... now I(a)=[ln(a+1)−ln( a+3)+ln(a+5)−...]+δ so answer is put a=3 [ln(4)−ln(6)+ln(8)−ln(10)...]+[ln3−ln5+ln7−ln11..] I=Σ_(r=1) ^∞ (−1)^(r−1) [ln(2+2r)+ln(1+2r)] =Σ_(r=1) ^∞ (−1)^(r−1) [ln2+ln{(1+r)(1+2r)}] pls others pls check...$
	$I (a) = \int_{0}^{1} \frac{x^{a} - 1}{1 + x^{2}} \times \frac{d x}{l n x}$ $\frac{{d I}_{}}{d a} = \int_{0}^{1} \frac{\partial}{\partial a} (\frac{x^{a} - 1}{1 + x^{2}}) \times \frac{d x}{l n x}$ $= \int_{0}^{1} \frac{x^{a} l n x}{1 + x^{2}} \times \frac{d x}{l n x}$ $= \int_{0}^{1} \frac{x^{a}}{1 + x^{2}} d x$ $= \int_{0}^{1} x^{a} (1 - x^{2} + x^{4} - x^{6} + . . .) d x$ $= \int_{0}^{1} x^{a} - x^{2 + a} + x^{4 + a} - x^{6 + a} + . . . d x$ $=∣ \frac{x^{a + 1}}{a + 1} - \frac{x^{a + 3}}{a + 3} - \frac{x^{a + 5}}{a + 5} - \frac{x^{a + 7}}{a + 7} . . . . ∣_{0}^{1}$ $= \frac{1}{a + 1} - \frac{1}{a + 3} + \frac{1}{a + 5} - \frac{1}{a + 7} + . . .$ $n o w \frac{d I}{d a} = \frac{1}{a + 1} - \frac{1}{a + 3} + \frac{1}{a + 5} - \frac{1}{a + 7} + . . .$ $\int d I = \int (\frac{1}{a + 1} - \frac{1}{a + 3} + \frac{1}{a + 5} - \frac{1}{a + 7} + . . .) d a$ $I (a) = [l n (a + 1) - l n (a + 3) + l n (a + 5) - . . .] + c$ $I (a) = \int_{0}^{1} \frac{x^{a} - 1}{(1 + x^{2}) l n x} d x$ $w h e n a = 0 I (a) = 0$ $s o 0 = [l n 1 - l n 3 + l n 5 - l n 7 . . .] + c$ $c = - l n 1 + l n 3 - l n 5 + l n 7 . . .$ $n o w I (a) = [l n (a + 1) - l n (a + 3) + l n (a + 5) - . . .] + δ$ $s o a n s w e r i s p u t a = 3$ $[l n (4) - l n (6) + l n (8) - l n (10) . . .] + [l n 3 - l n 5 + l n 7 - l n 11 . .]$ $I = \underset{r = 1}{\sum^{\infty}} {(- 1)}^{r - 1} [l n (2 + 2 r) + l n (1 + 2 r)]$ $= \underset{r = 1}{\sum^{\infty}} {(- 1)}^{r - 1} [l n 2 + l n {(1 + r) (1 + 2 r)}]$ $p l s o t h e r s p l s c h e c k . . .$
	Commented by Tawa1 last updated on 15/Jan/19

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