Blog: How do you solve the integration of (sqrt(x/(x-1))-1)/x dx Do you know? I tried substitution, but with out any result... :/

Friday, May 23, 2008

How do you solve the integration of (sqrt(x/(x-1))-1)/x dx Do you know? I tried substitution, but with out any result... :/

Hello!

As I understand, we are speaking about

$\displaystyle\int\frac{{\sqrt{{\frac{{x}}{{{x}-{1}}}}}-{1}}}{{x}}{\left.{d}{x}\right.}.$

We can omit the last term, -1/x, because we know its integral (ln|x|+C).

The remaining part is

$\displaystyle\int\frac{\sqrt{{\frac{{x}}{{{x}-{1}}}}}}{{x}}{\left.{d}{x}\right.}=\int\frac{{1}}{\sqrt{{{x}{\left({x}-{1}\right)}}}}{\left.{d}{x}\right.}.$

Make some transformations with the function under integral, and assume that $\displaystyle{x}\gt{1}:$

$\displaystyle\frac{{1}}{\sqrt{{{x}{\left({x}-{1}\right)}}}}=\frac{{1}}{\sqrt{{{x}{\left({x}-{1}\right)}}}}\cdot\frac{{\sqrt{{{x}}}+\sqrt{{{x}-{1}}}}}{{\sqrt{{{x}}}+\sqrt{{{x}-{1}}}}}=$

$\displaystyle=\frac{{\frac{{1}}{\sqrt{{{x}-{1}}}}+\frac{{1}}{\sqrt{{{x}}}}}}{{\sqrt{{{x}}}+\sqrt{{{x}-{1}}}}}=$

$\displaystyle={2}\cdot\frac{{{\left(\sqrt{{{x}}}+\sqrt{{{x}-{1}}}\right)}'}}{{\sqrt{{{x}}}+\sqrt{{{x}-{1}}}}}.$

Now we can integrate it, because $\displaystyle\int\frac{{1}}{{u}}{d}{u}={\ln}{\left|{u}\right|}+{C}:$

$\displaystyle\int\frac{{1}}{\sqrt{{{x}{\left({x}-{1}\right)}}}}{\left.{d}{x}\right.}={2}\cdot{\ln{{\left(\sqrt{{{x}}}+\sqrt{{{x}-{1}}}\right)}}}+{C}.$

This is the answer for x>0 and without omitted $\displaystyle-{\ln{{\left({x}\right)}}}.$

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Friday, May 23, 2008

How do you solve the integration of (sqrt(x/(x-1))-1)/x dx Do you know? I tried substitution, but with out any result... :/

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