∫_( 0) ^( 1) ((ln(1+x))/(1+x^2 ))dx


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Question Number 122963 by Lordose last updated on 21/Nov/20

$\int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} dx$
	Answered by Dwaipayan Shikari last updated on 21/Nov/20

	$\int_{0}^{\frac{π}{4}} \frac{l o g (1 + t a n θ)}{1 + {t a n}^{2} θ} . {s e c}^{2} θ d θ$ $= \int_{0}^{\frac{π}{4}} l o g (\sqrt{2} (c o s (\frac{π}{4} - x)) - l o g (c o s x) d x$ $= \int_{0}^{\frac{π}{4}} \frac{1}{2} l o g (2) + \int_{0}^{\frac{π}{4}} l o g (c o s (\frac{π}{4} - x)) - l o g (c o s x)$ $= \frac{π}{8} l o g (2) + I$ $I = \int_{0}^{\frac{π}{4}} l o g (c o s (\frac{π}{4} - x)) - l o g (c o s x) = \int_{0}^{\frac{π}{4}} l o g (c o s x) - l o g (c o s (\frac{π}{4} - x))$ $I = 0$ $S o a n s w e r i s$ $\Rightarrow \frac{π}{8} l o g (2)$
	Answered by mnjuly1970 last updated on 21/Nov/20
	$solution: x=tan(y)⇒ dx= (1+tan^2 (y))dy I =∫_0 ^( 1) ((ln(1+x))/(1+x^2 ))dx=∫_0 ^( (π/4)) ln(1+tan(y))dy =∫_0 ^( (π/4)) ln[1+tan((π/4)−y)]dy =∫_0 ^( (π/4)) ln(1+((1−tan(y))/(1+tan(y))))dy =∫_0 ^( (π/4)) {ln(2)−ln(1+tan(y))}dy =(π/4)ln(2)−I 2I=(π/4)ln(2) ⇒I=(π/8)ln(2) ✓$
	$s o l u t i o n :$ $x = t a n (y) \Rightarrow d x = (1 + {t a n}^{2} (y)) d y$ $I = \int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x = \int_{0}^{\frac{π}{4}} l n (1 + t a n (y)) d y$ $= \int_{0}^{\frac{π}{4}} l n [1 + t a n (\frac{π}{4} - y)] d y$ $= \int_{0}^{\frac{π}{4}} l n (1 + \frac{1 - t a n (y)}{1 + t a n (y)}) d y$ $= \int_{0}^{\frac{π}{4}} {l n (2) - l n (1 + t a n (y))} d y$ $= \frac{π}{4} l n (2) - I$ $2 I = \frac{π}{4} l n (2) \Rightarrow I = \frac{π}{8} l n (2) ✓$
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