calculate ∫_0 ^1 ((ln(1+x))/((x+1)^2 ))dx


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Question Number 91271 by mathmax by abdo last updated on 29/Apr/20

$c a l c u l a t e \int_{0}^{1} \frac{l n (1 + x)}{{(x + 1)}^{2}} d x$
Commented by mathmax by abdo last updated on 29/Apr/20

$I = \int_{0}^{1} \frac{l n (1 + x)}{{(x + 1)}^{2}} d x b y p s r t s w e g e t$ $I = {[- \frac{1}{x + 1} l n (x + 1)]}_{0}^{1} + \int_{0}^{1} \frac{l n (x + 1)}{x + 1} d x$ $= - \frac{1}{2} l n (2) + \int_{1}^{2} \frac{l n t}{t} d t (x + 1 = t)$ $\int_{1}^{2} \frac{l n (t)}{t} d t =_{l n t = u} \int_{0}^{l n (2)} \frac{u}{e^{u}} e^{u} d u = \int_{0}^{l n (2)} u d u = {[\frac{u^{2}}{2}]}_{0}^{l n (2)}$ $= \frac{1}{2} {l n}^{2} (2) \Rightarrow I = - \frac{1}{2} l n (2) + \frac{1}{2} {l n}^{2} (2) .$
	Answered by M±th+et+s last updated on 29/Apr/20

	$I = \int_{0}^{1} \frac{l n (1 + x)}{x^{2} + 1} d x$ $p u t x = t a n (u) d x = {s e c}^{2} (u)$ $= \int_{0}^{\frac{π}{4}} \frac{l n (t a n (u) + 1)}{{t a n}^{2} u + 1} {s e c}^{2} (u) d u$ $= \int_{0}^{\frac{π}{4}} l n (t a n (u) + 1) d u$ $= \int_{0}^{\frac{π}{4}} l n (\frac{s i n u + c o s u}{c o s u}) d u$ $\int_{0}^{\frac{π}{4}} l n (s i n u + c o s u) - l n (c o s u) d u$ $\int_{0}^{\frac{π}{4}} l n (\sqrt{2} (c o s (\frac{π}{4} - u)) - l n (c o s u) d u$ $\int_{0}^{\frac{π}{4}} l n (\sqrt{2}) + l n (c o s (\frac{π}{4} - u)) - l n (c o s u) d u$ $J = \int_{0}^{\frac{π}{4}} l n (c o s (\frac{π}{4} - u)) d u$ $\frac{π}{4} - u = t d u = - d t$ $t h e n J = \int_{\frac{π}{4}}^{0} - l n (c o s (t)) d t \int_{0}^{\frac{π}{4}} l n (c o s (t)) d t$ $\int_{0}^{\frac{π}{4}} l n (c o s (t)) d t = \int_{0}^{\frac{π}{4}} l n (c o s (u)) d u$ $t h e n \int_{0}^{\frac{π}{4}} l n (\sqrt{2}) + l n (c o s (u)) - l n (c o s (u)) d u$ $= \int_{0}^{\frac{π}{4}} l n (\sqrt{2}) d u = l n \sqrt{2} \frac{π}{4} = \frac{π}{8} l n (2)$
	Commented by mathmax by abdo last updated on 29/Apr/20

	$t h a n k y o u s i r .$
	Commented by M±th+et+s last updated on 29/Apr/20

	$y o u a r e w e l c o m e$
	Commented by mathmax by abdo last updated on 29/Apr/20

	$l o o k s i r t h e i n t e g r a l i s \int_{0}^{1} \frac{l n (1 + x)}{{(1 + x)}^{2}} d x n o t \int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x$ $a n y w a y t h a n k f o r y o u r w o r k . . .$
	Commented by M±th+et+s last updated on 29/Apr/20

	$s o r r y, i {d i d n}^{'} t n o t i c e$
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