Nice-Calculus-I-Evaluate-I-0-ln-2-x-e-x-2e-x-2-dx-

Question Number 141560 by mnjuly1970 last updated on 20/May/21

\dots \dots . . N i c e \dots . C a l c u l u s (I) \dots \dots

E v a l u a t e ::

I := \int_{0}^{l n (2)} \frac{x}{e^{x} + 2 e^{- x} - 2} d x = ?

\dots \dots

Answered by mindispower last updated on 20/May/21

= \int_{0}^{l n (2)} \frac{{x e}^{x}}{{(e^{x} - 1)}^{2} + 1} d x

u = e^{x} - 1

= \int_{0}^{1} \frac{l n (1 + u)}{u^{2} + 1} d u, u = t g (t)

= \int_{0}^{\frac{π}{4}} l n (1 + t g (t))

= \int_{0}^{\frac{π}{4}} l n (\frac{\sqrt{2} s i n (\frac{π}{4} + u)}{c o s (u)}) d u

= \frac{π}{4} l n (\sqrt{2}) + \int_{0}^{\frac{π}{4}} l n (s i n (\frac{π}{4} + u)) d u - \int_{0}^{\frac{π}{4}} l n (c o s (u)) d u

\int_{0}^{\frac{π}{4}} l n (s i n (u + \frac{π}{4})) d u = \int_{0}^{\frac{π}{4}} l n (c o s (u)) d u ∣ u \to \frac{π}{4} - u

\Leftrightarrow \int_{0}^{l n (2)} \frac{x}{e^{x} + 2 e^{- x} - 2} d x = \frac{π}{4} l n (\sqrt{2}) = \frac{π l n (2)}{8}

Commented by mnjuly1970 last updated on 20/May/21

g r a t e f u l s i r p o w e r \dots

Commented by mindispower last updated on 20/May/21

p l e a s u r s i r