How do you find $\int \frac{x - 1}{\sqrt{x^{2} - 2 x}}$ $int (x-1)/sqrt(x^2-2x)$ ?

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How do you find $\int \frac{x - 1}{\sqrt{x^{2} - 2 x}}$ $int (x-1)/sqrt(x^2-2x)$ ?

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Answer 1 · 2016-11-10T20:45:53

$\sqrt{x^{2} - 2 x} + C$ $sqrt(x^2-2x)+C$

Explanation:

$I = \int \frac{x - 1}{\sqrt{x^{2} - 2 x}} d x = \frac{1}{2} \int \frac{2 x - 2}{\sqrt{x^{2} - 2 x}} d x$ $I=int(x-1)/sqrt(x^2-2x)dx=1/2int(2x-2)/sqrt(x^2-2x)dx$

Substituting $u = x^{2} - 2 x$ $u=x^2-2x$ yields $d u = (2 x - 2) d x$ $du=(2x-2)dx$ .

$I = \frac{1}{2} \int \frac{d u}{\sqrt{u}} = \frac{1}{2} \int u^{- \frac{1}{2}} d u$ $I=1/2int(du)/sqrtu=1/2intu^(-1/2)du$

With the rule $\int u^{n} d u = \frac{u^{n + 1}}{n + 1} + C$ $intu^ndu=u^(n+1)/(n+1)+C$ , we see that

$I = \frac{1}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C = u^{\frac{1}{2}} + C = \sqrt{u} + C = \sqrt{x^{2} - 2 x} + C$ $I=1/2u^(1/2)/(1/2)+C=u^(1/2)+C=sqrtu+C=sqrt(x^2-2x)+C$

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How do you find $\int \frac{x - 1}{\sqrt{x^{2} - 2 x}}$ $int (x-1)/sqrt(x^2-2x)$ ?

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How do you find int (x-1)/sqrt(x^2-2x)∫x−1√x2−2xint (x-1)/sqrt(x^2-2x)?

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How do you find $\int \frac{x - 1}{\sqrt{x^{2} - 2 x}}$ $int (x-1)/sqrt(x^2-2x)$ ?