# Thread: [SOLVED] Evaluate the Integral

1. ## [SOLVED] Evaluate the Integral

I really need help figuring out this integral:
$\displaystyle integral 5x/sqrt(5-x^4)$

2. Sorry about the format of the question, I don't really know how to get the symbols in

3. Originally Posted by miniwheats
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... $\displaystyle u = x^2$

Originally Posted by miniwheats
Sorry about the format of the question, I don't really know how to get the symbols in
see here

4. Thanks a lot for your help!

5. Originally Posted by miniwheats
Thanks a lot for your help!
you're welcome

6. 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?

7. Originally Posted by miniwheats
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:

= $\displaystyle \frac{3}{2}\int\frac{du}{1+u^2}$

= $\displaystyle \frac{3}{2}\cdot {\color{red} \frac{1}{u}\arctan(\frac{1}{u})}$

= $\displaystyle \frac{3}{2}[\frac{1}{e^{2t}}\arctan(\frac{1}{e^{2t}})]$

which isn't the right answer, where did I go wrong?
what is in red is your mistake. that formula is wrong. heck, you did not need that formula (the one you were thinking about).

$\displaystyle \int \frac {dx}{1 + x^2} = \arctan x + C$

8. Ack, my mistake. I made that question more complicated than it had to be by forgetting that $\displaystyle \int\frac{dx}{1+x^2} = arctan(x) + C$ -_-