find-0-1-x-3-4-1-x-1-4-x-dx-

Question Number 114395 by mathmax by abdo last updated on 18/Sep/20

find \int_{0}^{\infty} \frac{{(1 + x)}^{- \frac{3}{4}} - {(1 + x)}^{- \frac{1}{4}}}{x} dx

Answered by Olaf last updated on 19/Sep/20

I = \int_{0}^{\infty} \frac{{(1 + x)}^{- \frac{3}{4}} - {(1 + x)}^{- \frac{1}{4}}}{x} d x

u = {(1 + x)}^{\frac{1}{4}}

d u = \frac{1}{4} {(1 + x)}^{- \frac{3}{4}} d x = \frac{d x}{4 u^{3}}

I = \int_{1}^{\infty} \frac{\frac{1}{u^{3}} - \frac{1}{u}}{u^{4} - 1} . 4 u^{3} d u

I = 4 \int_{1}^{\infty} \frac{1 - u^{2}}{u^{4} - 1} d u

I = - 4 \int_{1}^{\infty} \frac{d u}{1 + u^{2}}

I = - 4 {[\arctan u]}_{1}^{\infty} = - 4 (\frac{π}{2} - \frac{π}{4}) = - π

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