calculate ∫_(−1) ^1 (((x^4 +x^2 +1)^2 +e^x )/(e^x +1))dx


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Question Number 57421 by Abdo msup. last updated on 03/Apr/19

$c a l c u l a t e \int_{- 1}^{1} \frac{{(x^{4} + x^{2} + 1)}^{2} + e^{x}}{e^{x} + 1} d x$
	Answered by einsteindrmaths@hotmail.fr last updated on 04/Apr/19

	$= \int_{- 1} \overset{1}{} \frac{{(x^{4} + x^{2} + 1)}^{2} + e^{x}}{e^{x} + 1} d x = \int_{- 1} \overset{1}{} \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{x} + 1} d x + \underset{- 1}{\int^{1}} \frac{e^{x}}{e^{x} + 1} d x$ $t h e s e c o n d e i n t e g r a l i s e a s y t o e v a l u a t$ $\int \frac{e^{x}}{e^{x} + 1} d x = l n (e^{x} + 1) + c$ $l e t s f i n d \underset{- 1}{\int^{1}} \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{x} + 1} d x = \underset{- 1}{\int^{0}} \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{x} + 1} d x + \underset{0}{\int^{1}} \frac{{(x^{4} + x^{2} + 1)}^{4}}{e^{x} + 1} d x$ $w e p u t y = - x i n t h e f i r s t e w e g e t \int_{0}^{1} \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{- y} + 1} d y + \int_{0}^{1} \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{x} + 1} d x$ $= \underset{0}{\int^{1}} [\frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{x} + 1} + \frac{{(x^{4} + x^{2} + 1)}^{2}}{e^{- x} + 1}] d x = \int_{0}^{1} {(x^{4} + x^{2} + 1)}^{2} d x = \underset{0}{\int^{1}} (x^{8} + 2 x^{6} + 3 x^{4} + 2 x^{2} + 1) d x$ $e a s y t o e v a l u a t$
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