1. ## Integration

How do I even go about solving this? I get as far as u = x^4+1 and du = 4x^3

2. what if you put $\displaystyle t=x^2$ ?

3. Alright so I changed it so that it's $\displaystyle u = x^2+1$ and $\displaystyle du = 2x dx$.

Eventually I get to $\displaystyle 4 \int du * 1/u^2$ and couldn't proceed.

The du becomes x^2 but I don't know what to do with the second one. It could become arctan u, but the fact that it's $\displaystyle 1/u^2$ and not just $\displaystyle 1/u$ prevents that from happening.

4. no, actually your substitution won't work since you need to express $\displaystyle x^4$ in terms of your substitution and that is nasty.

try my sub. and see what happens.

5. Originally Posted by Krizalid
what if you put $\displaystyle t=x^2$ ?
What's t? du or u?

6. it's any variable, that makes no difference.

7. Alright so

$\displaystyle 4 \int 2x/(x^4+1)$
$\displaystyle u = x^2$, $\displaystyle du = 2x$
$\displaystyle 4 \int du/(u^2+1)$

Do I separate? Does it become $\displaystyle 4 \int du * 1/(u^2+1)$?

In which case the answer should be 4 * x^2 * arctan x^2 + C, right?

8. no, it's not the answer, but it's close.

Once you get the integral to $\displaystyle 4\int{\frac{1}{u^2 + 1}\,du}$ use the same trigonometric substitution I explained to you in another thread.