integral of (x^2)/root(4-x)dx
integral of e^(4x)/(e^2x-4)^2
Thanks in advance. ^_^
$\displaystyle \int\frac{e^{4x}}{(e^{2x}-4)^{2}}dx$
You could let $\displaystyle u=e^{2x}, \;\ du=2e^{2x}dx, \;\ \frac{du}{2}=e^{2x}dx$
Make the subs and get:
$\displaystyle \frac{1}{2}\int\frac{u}{(u-4)^{2}}du=\frac{1}{2}\left[\int\frac{1}{u-4}du+\int\frac{4}{(u-4)^{2}}du\right]$
These are relatively easy integrals to work with now.
You can also rearrange it in other ways by using substitutions.
Let $\displaystyle u = 4 - x \implies du = -dx$
$\displaystyle \int \frac{x^2}{\sqrt{4 - x}}~dx = \int \frac{(4 - u)^2}{\sqrt{u}}~-du$
$\displaystyle = -\int\frac{4 - 4u + u^2}{u^{1/2}}~du$
$\displaystyle = -\int 4~u^{-1/2}~du + \int 4u^{1/2}~du - \int u^{3/2}~du$
which you should be able to do easily.
-Dan