I really need help figuring out this integral:
$\displaystyle integral 5x/sqrt(5-x^4)$
you can use a trig substitution: $\displaystyle x^2 = \sqrt{5} \sin \theta$
if you don't like the x squared thing, you can do a regular substitution first... $\displaystyle u = x^2$
see here
Sorry, I just have one more question.
$\displaystyle \int\frac{3e^{2t} dt}{1+e^{4t}}$
I substituted $\displaystyle u= e^{2t}$, $\displaystyle du=2e^{2t}dt $ to get:
= $\displaystyle \frac{3}{2}\int\frac{du}{1+u^2}$
= $\displaystyle \frac{3}{2}\cdot\frac{1}{u}\arctan(\frac{1}{u})$
= $\displaystyle \frac{3}{2}[\frac{1}{e^{2t}}\arctan(\frac{1}{e^{2t}})]$
which isn't the right answer, where did I go wrong?