# Evaluating integrals with radicals and fractions

• Aug 10th 2010, 08:37 PM
bobsanchez
Evaluating integrals with radicals and fractions
Two integration problems are giving me a lot of trouble.

(2x + 2)/(x^2 + 2x)^(1/2)

and

dx/(x^1/2)(1 - x^1/2)^4

I can't get the substitution technique to work out on these two for some reason.
• Aug 10th 2010, 08:42 PM
Prove It
$\displaystyle \int{\frac{2x+2}{(x^2 + 2x)^{\frac{1}{2}}}\,dx} = \int{(2x+2)(x^2+2x)^{-\frac{1}{2}}\,dx}$.

Let $\displaystyle u = x^2+2x$ so that $\displaystyle du = (2x+2)dx$ and the integral becomes

$\displaystyle \int{u^{-\frac{1}{2}}\,du}$

$\displaystyle = \frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C$

$\displaystyle = 2u^{\frac{1}{2}}+C$

$\displaystyle = 2(x^2+2x)^{\frac{1}{2}}+C$.
• Aug 10th 2010, 08:49 PM
Prove It
$\displaystyle \int{\frac{dx}{x^{\frac{1}{2}}(1-x^{\frac{1}{2}})^4}} = \int{x^{-\frac{1}{2}}(1-x^{\frac{1}{2}})^{-4}\,dx}$

$\displaystyle = -2\int{-\frac{1}{2}x^{-\frac{1}{2}}(1-x^{\frac{1}{2}})^{-4}\,dx}$.

Let $\displaystyle u=1-x^{\frac{1}{2}}$ so that $\displaystyle du = -\frac{1}{2}x^{-\frac{1}{2}}\,dx$ and the integral becomes

$\displaystyle -2\int{u^{-4}\,du}$

$\displaystyle = -2\left(\frac{u^{-3}}{-3}\right) + C$

$\displaystyle = \frac{2}{3u^3} + C$

$\displaystyle = \frac{2}{3(1-x^{\frac{1}{2}})^3} + C$.