find ∫_0 ^1 ((ln(1+x^2 ))/(1+x))dx


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Question Number 112446 by mathmax by abdo last updated on 07/Sep/20

$find \int_{0}^{1} \frac{\ln (1 + x^{2})}{1 + x} dx$
	Answered by mathdave last updated on 08/Sep/20

	$s o l u t i o n t o$ $l e t I = \int_{0}^{1} \frac{\ln (1 + x^{2})}{1 + x} d x u s i n g I B P$ $I =∣ \ln (1 + x) \ln (1 + x^{2}) ∣_{0}^{1} - 2 \int_{0}^{1} \frac{x \ln (1 + x)}{1 + x^{2}} d x$ $I = \ln^{2} (2) - 2 I_{1} . . . . . . . . (1)$ $L e t$ $I_{1} = \int_{0}^{1} \frac{x \ln (1 + x)}{1 + x^{2}} d x = \int_{0}^{1} [\frac{(1 + x^{2} - 1)}{1 + x^{2}}] \frac{\ln (1 + x)}{x} d x$ $I_{1} = \int_{0}^{1} \frac{\ln (1 + x)}{x} d x - \int_{0}^{1} \frac{\ln (1 + x)}{x (1 + x^{2})} d x$ $a d o p t f a y n m a n n t r i c k$ $I_{1} = \int_{0}^{1} \frac{\ln (1 + x)}{x} d x - \int_{0}^{1} \frac{1}{1 + x^{2}} (\int_{0}^{1} \frac{1}{1 + x y} d y) d x$ $I_{1} = η (2) - \int_{0}^{1} \frac{1}{1 + y^{2}} \int_{0}^{1} [\frac{y^{2}}{1 + x y} + \frac{1 - x y}{1 + x^{2}}] d x d y$ $I_{1} = \frac{π^{2}}{12} - \int_{0}^{1} \frac{1}{1 + y^{2}} [y \ln (1 + y) + \frac{π}{4} - \frac{y}{2} \ln 2] d y$ $I_{1} = \frac{π^{2}}{12} - \int_{0}^{1} \frac{y \ln (1 + y)}{1 + y^{2}} - \frac{π}{4} \int_{0}^{1} \frac{1}{1 + y^{2}} d y + \frac{1}{2} \ln 2 \int_{0}^{1} \frac{y}{1 + y^{2}} d y$ $I_{1} = \frac{π^{2}}{12} - I_{1} - \frac{π^{2}}{16} + \frac{\ln^{2} (2)}{4}$ $2 I_{1} = \frac{π^{2}}{48} + \frac{\ln^{2} (2)}{4}$ $∵ I_{1} = \frac{π^{2}}{96} + \frac{\ln^{2} (2)}{8}$ $r e c a l l t h a t I = \ln^{2} (2) - 2 I_{1} . . . . . . . . . (1)$ $∵ I = \ln^{2} (2) - 2 [\frac{π^{2}}{96} + \frac{\ln^{2} (2)}{8}]$ $I = \ln^{2} (2) - \frac{. π^{2}}{48} - \frac{\ln^{2} (2)}{4}$ $I = \frac{3}{4} \ln^{2} (2) - \frac{π^{2}}{48}$ $∵ \int_{0}^{1} \frac{\ln (1 + x^{2})}{1 + x} d x = \frac{3}{4} \ln^{2} (2) - \frac{π^{2}}{48}$ $b y m a t h d a v e (08 / 09 / 2020)$
	Commented by mathmax by abdo last updated on 08/Sep/20

	$thanks$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
	Answered by mnjuly1970 last updated on 08/Sep/20

	Commented by mathmax by abdo last updated on 08/Sep/20

	$thanks$
	Commented by mnjuly1970 last updated on 08/Sep/20

	$g r a t e f u l s i r . t o m e$ $y o u a r e^{'} {o s t a d}^{'} (i n o u r$ $l a n g u a g e m e a n s^{'} m a s t e r$ $a n d {s c h o l a r}^{'} i n h i s f i e l d)$ $i n m a t h e m a t i c s . t h a n k$ $y o u s o m u c h s i r f o r y o u r$ $e f f o r t . . . .$
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