∫_(−1) ^1 x cosh(x) ln(1+e^x ) dx


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Question Number 89846 by M±th+et£s last updated on 19/Apr/20

$\int_{- 1}^{1} x c o s h (x) l n (1 + e^{x}) d x$
Commented by M±th+et£s last updated on 19/Apr/20

$t h a n k y o u v e r r y m u c h$
Commented by abdomathmax last updated on 19/Apr/20

$I = \int_{- 1}^{1} x \frac{e^{x} + e^{- x}}{2} l n (1 + e^{x}) d t$ $= \frac{1}{2} \int_{- 1}^{1} {x e}^{x} l n (1 + e^{x}) d x + \frac{1}{2} \int_{- 1}^{1} x e^{- x} l n (1 + e^{x}) d x$ $= H + K$ $H = \int_{- 1}^{1} {x e}^{x} l n (1 + e^{x}) d x b y p a r t s u^{^{'}} = {x e}^{x} a n d$ $v = l n (1 + e^{x}) \Rightarrow u = \int {x e}^{x} = {x e}^{x} - \int e^{x} = (x - 1) e^{x}$ $H = {[(x - 1) e^{x} l n (1 + e^{x})]}_{- 1}^{1} - \int_{- 1}^{1} (x - 1) e^{x} \times \frac{e^{x}}{1 + e^{x}} d x$ $= 2 e^{- 1} l n (1 + e^{- 1}) - \int_{- 1}^{1} \frac{(x - 1) e^{2 x}}{1 + e^{x}} d x$ $\int_{- 1}^{1} \frac{(x - 1) e^{2 x}}{1 + e^{x}} d x =_{e^{x} = t} \int_{e^{- 1}}^{e} \frac{(l n t - 1) t^{2}}{1 + t} \times \frac{d t}{t}$ $= \int_{e^{- 1}}^{e} \frac{t l n t - t}{1 + t} d t = {[l n (t) l n (t)]}_{e^{- 1}}^{e} - \int_{e^{- 1}}^{e} l n t l n (1 + t) d t$ $= 2 - \int_{e^{- 1}}^{e} l n (t) l n (1 + t) d t w e h a v e$ $\int_{e^{- 1}}^{e} l n (t) l n (1 + t) d t = \int_{e^{- 1}}^{1} l n (t) l n (1 + t) d t +$ $\int_{1}^{e} l n (t) l n (1 + t) d t$ ${l n}^{^{'}} (1 + t) = \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n} \Rightarrow$ $l n (1 + t) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} t^{n + 1} + c (c = 0)$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} t^{n}}{n} \Rightarrow$ $\int_{e^{- 1}}^{1} l n (t) l n (1 + t) d t = \int_{e^{- 1}}^{1} l n t (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} t^{n}) d t$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{e^{- 1}}^{1} t^{n} l n (t) d t$ $A_{n} = \int_{e^{- 1}}^{1} t^{n} l n (t) d t = {[\frac{t^{n + 1}}{n + 1} l n (t)]}_{e^{- 1}}^{1} - \int_{e^{- 1}}^{1} \frac{t^{n}}{n + 1} d t$ $= \frac{1}{n + 1} e^{- n - 1} - \frac{1}{n + 1} {[\frac{1}{n + 1} t^{n + 1}]}_{e^{- 1}}^{1}$ $. . . b e c o n t i n u e d . . .$
Commented by mathmax by abdo last updated on 19/Apr/20

$y o u a r e w e l c o m e .$
	Answered by M±th+et£s last updated on 19/Apr/20

	$\int_{- 1}^{1} x c o s h (x) l n (1 + e^{x}) . . . . . . . (1)$ $l e t y = - x \Rightarrow d y = - d x$ $I = \int_{- 1}^{1} - y c o s h (- y) l n (1 + e^{y}) d x$ $I = \int_{- 1}^{1} - y c o s h (- y) l n (\frac{1 + e^{y}}{e^{y}}) d y$ $I = \int_{- 1}^{1} - y c o s h (y) l n (1 + e^{y}) + y c o s h (y) l n (e^{y}) d y$ $I = \int_{- 1}^{1} - y c o s h (y) l n (1 + e^{y}) + y^{2} c o s h (y) d y . . . . (2)$ $a d d (1) t o (2)$ $2 I = \int_{- 1}^{1} x^{2} c o s h (x) d x$ $i n t e g r a t e b y p a r t s$ $2 I = (x^{2} + 2) s i n h (x) - 2 x c o s h (x) ∣_{- 1}^{1} ._{}^{}$ $2 I = 6 s i n h (1) - 4 c o s h (1)$ $I = 3 s i n h (1) - 2 c o s h (1)$
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