∫_( 1) ^(2)^(1/7) (1/(x(2x^7 + 1))) dx =


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Question Number 79452 by Vishal Sharma last updated on 25/Jan/20

$\underset{1}{\int^{\sqrt[7]{2}}} \frac{1}{x (2 x^{7} + 1)} d x =$
Commented by john santu last updated on 25/Jan/20
$look ∫ (dx/(2x^8 +x))=∫ (dx/(x^8 (2+(1/x^7 )))) let : u = 2+(1/x^7 ) ⇒du = −(7/x^8 )dx (dx/x^8 ) = −(du/7) ∫ ((−du)/(7u)) = −(1/7)ln (2+(1/x^7 ))∣_1 ^(2)^(1/(7 )) = −(1/7){ln(5/2)−ln 3}=ln((6/5))^(1/(7 ))$
$l o o k \int \frac{d x}{2 x^{8} + x} = \int \frac{d x}{x^{8} (2 + \frac{1}{x^{7}})}$ $l e t : u = 2 + \frac{1}{x^{7}} \Rightarrow d u = - \frac{7}{x^{8}} d x$ $\frac{d x}{x^{8}} = - \frac{d u}{7}$ $\int \frac{- d u}{7 u} = - \frac{1}{7} l n (2 + \frac{1}{x^{7}}) \underset{1}{\overset{\sqrt[7]{2}}{∣}}$ $= - \frac{1}{7} {l n \frac{5}{2} - l n 3} = l n \sqrt[7]{\frac{6}{5}}$
	Answered by mind is power last updated on 25/Jan/20

	$= \int \frac{x^{6}}{x^{7} (2 x^{7} + 1)} d x$ $p u t y = x^{7}$
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