dx-1-x-3-

Question Number 129855 by liberty last updated on 20/Jan/21

ϑ = \int \frac{d x}{{(1 + \sqrt{x})}^{3}}

Answered by EDWIN88 last updated on 20/Jan/21

ϑ = \int \frac{dx}{{(\sqrt{x})}^{3} {(1 + x^{- 1 / 2})}^{3}} = \int \frac{x^{- 3 / 2}}{{(1 + x^{- 1 / 2})}^{3}} dx

let φ = 1 + x^{- 1 / 2} \to d φ = - \frac{1}{2} x^{- 3 / 2} dx

ϑ = - 2 \int \frac{d φ}{φ^{3}} = - 2 \int φ^{- 3} d φ

φ = \frac{1}{φ^{2}} + C = \frac{1}{{(\frac{\sqrt{x} + 1}{\sqrt{x}})}^{2}} + C

φ = \frac{x}{{(\sqrt{x} + 1)}^{2}} + C

Answered by stelor last updated on 20/Jan/21

let u = \sqrt{x} s o, d u = \frac{1}{2 \sqrt{x}} d x = d u = \frac{1}{2 u} d x

v = \int \frac{2 udu}{{(1 + u)}^{3}} w h e r e u = \sqrt{x}

l e t t = 1 + u s o, du = dt

v = 2 \int \frac{(t - 1)}{t^{3}} d t w h e r e t = u - 1 = \sqrt{x} - 1

v = 2 [\int (\frac{1}{t^{2}} - \frac{1}{t^{3}}) d t] = 2 (- \frac{1}{t} + \frac{1}{2 t^{2}} + c)

v = 2 (- \frac{1}{\sqrt{x} - 1} + \frac{1}{2 {(\sqrt{x} - 1)}^{2}} + c)

v = \frac{3 - 2 \sqrt{x}}{{(\sqrt{x} - 1)}^{2}} + c

Commented by bemath last updated on 20/Jan/21

t = 1 + u but t = u - 1; ambiguous

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