How do you evaluate the integral $int x/(sqrt(x-1)-sqrtx)$ ?

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Answer 1 · 2017-01-29T00:51:51

$-2/5(x-1)^(5/2)-2/3(x-1)^(3/2)-2/5x^(5/2)+C$

Multiply first by the conjugate of the denominator to simplify the integrand.

$I=int(x(sqrt(x-1)+sqrtx))/((sqrt(x-1)-sqrtx)(sqrt(x-1)+sqrtx))dx$

The denominator is now in the form $(a-b)(a+b)=a^2-b^2$ , where $a=sqrt(x-1)$ and $b=x$ .

$I=int(x(sqrt(x-1)+sqrtx))/((x-1)-x)dx$

$I=-intx(sqrt(x-1)+sqrtx)dx$

Distributing, splitting up the integral and rewriting:

$I=-intx(x-1)^(1/2)dx-intx^(3/2)dx$

The second integral can be directly integrated using $intx^ndx=x^(n+1)/(n+1)+C$ , where $n!=-1$ .

$I=-intx(x-1)^(1/2)dx-x^(5/2)/(5/2)$

$I=-intx(x-1)^(1/2)dx-2/5x^(5/2)$

For the remaining integral, let $u=x-1$ . This implies that $x=u+1$ and $du=dx$ .

$I=-int(u+1)u^(1/2)du-2/5x^(5/2)$

Distributing $(u+1)u^(1/2)$ into separate integrals:

$I=-intu^(3/2)du-intu^(1/2)du-2/5x^(5/2)$

Using the rule from earlier:

$I=-u^(5/2)/(5/2)-u^(3/2)/(3/2)-2/5x^(5/2)+C$

From $u=x-1$ :

$I=-2/5(x-1)^(5/2)-2/3(x-1)^(3/2)-2/5x^(5/2)+C$

How do you evaluate the integral int x/(sqrt(x-1)-sqrtx)int x/(sqrt(x-1)-sqrtx)?