1-1-x-dx-

Question Number 175191 by ajfour last updated on 22/Aug/22

\int \sqrt{1 - \frac{1}{x}} d x = ?

Commented by ajfour last updated on 22/Aug/22

T h a n k y o u b o t h s i r s .

Answered by Ar Brandon last updated on 22/Aug/22

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

= \frac{1}{2} \int \frac{2 x - 1}{\sqrt{x^{2} - x}} d x - \frac{1}{2} \int \frac{1}{\sqrt{{(x - \frac{1}{2})}^{2} - \frac{1}{4}}} d x

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

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

Answered by MJS_new last updated on 22/Aug/22

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

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

= - 2 \int \frac{d t}{{(t^{2} - 1)}^{2}} = \frac{t}{t^{2} - 1} + \frac{1}{2} \ln \frac{t - 1}{t + 1} =

= \sqrt{x (x - 1)} + \ln (\sqrt{x} - \sqrt{x - 1}) + C

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