Question-168857

Question Number 168857 by Sotoberry last updated on 19/Apr/22

Answered by MJS_new last updated on 20/Apr/22

\int \frac{1}{x^{3}} \sqrt{\frac{x^{2} - 1}{x + 1}} d x = \int \frac{\sqrt{x - 1}}{x^{3}} d x =

[t = \sqrt{x - 1} \to d x = 2 \sqrt{x - 1} d t]

= 2 \int \frac{t^{2}}{{(t^{2} + 1)}^{3}} d t = \frac{t (t^{2} - 1)}{4 {(t^{2} + 1)}^{2}} + \frac{1}{4} \arctan t =

= \frac{(x - 2) \sqrt{x - 1}}{4 x^{2}} + \frac{1}{4} \arctan \sqrt{x - 1} + C

Answered by MJS_new last updated on 20/Apr/22

\int \sqrt{\frac{x - 1}{x^{5}}} d x =

t = \sqrt{\frac{x - 1}{x}} \to d x = 2 x^{2} \sqrt{\frac{x - 1}{x}} d t]

= 2 \int t^{2} d t = \frac{2 t^{3}}{3} = \frac{2}{3} \sqrt{\frac{{(x - 1)}^{3}}{x^{3}}} + C

Answered by cortano1 last updated on 20/Apr/22

\int \frac{\sqrt{x - 1}}{\sqrt{x^{5}}} d x = \int \frac{\sqrt{1 - x^{- 1}} d x}{x^{2}}

= \int x^{- 2} \sqrt{1 - x^{- 1}} d x

[l e t u^{2} = 1 - x^{- 1} \Rightarrow 2 u d u = x^{- 2} d x]

I = \int u (2 u d u) = \frac{2}{3} u^{3} + c

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

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