Question

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

Answer 1

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

Explanation:

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`