# Math Help - Integration Problem

1. ## Integration Problem

$\displaystyle \int \frac{2 x^2 + 1}{x^2}\, dx$

What technique do I use to figure this out? I know the answer, but I can't figure out how to arrive at it!

I am trying method of substitution:
$u=2x^2+1$
$du=4xdx$

I don't see any possible substitutions...
I then tried:
$x=\sqrt{\frac{u-1}{2}}$

when I put this back into the original I have:
$\frac{2u*du}{u-1}$

now, I am not sure where to go from here.. Am I on the right track?

Thanks!

2. $\displaystyle\frac{2x^2+1}{x^2}=2+\frac{1}{x^2}$

3. Originally Posted by Vamz
$\displaystyle \int \frac{2 x^2 + 1}{x^2}\, dx$

What technique do I use to figure this out? I know the answer, but I can't figure out how to arrive at it!

I am trying method of substitution:
$u=2x^2+1$
$du=4xdx$

I don't see any possible substitutions...
I then tried:
$x=\sqrt{\frac{u-1}{2}}$

when I put this back into the original I have:
$\frac{2u*du}{u-1}$

now, I am not sure where to go from here.. Am I on the right track?

Thanks!
$\displaystyle \frac{2x^2+1}{x^2} = 2 + x^{-2}$

now integrate ...

4. Thank you very much I got it.

I have one more question:

$\int x^5e^{x^6}$

I was able to solve this by method of staring & got $\frac{e^{x^6}}{6}$

I tried to apply the method of substitution to this for kicks, and heres what happens:

$u=e^{x^6}$
$du=6x^5e^{x^6}dx$

As a general rule for the future, what exactly does it mean when my du contains my u? What should I do in these scenarios?

$
\int \frac{du}{6} = \frac{1}{6}\int du
$

Would this be the proper way to proceed?
When all I have is my du in my expression, does that always mean to stop "integrating" and to just write it as-is?

Thanks!

5. Originally Posted by Vamz
Thank you very much I got it.

I have one more question:

$\int x^5e^{x^6}$

I was able to solve this by method of staring & got $\frac{e^{x^6}}{6}$

I tried to apply the method of substitution to this for kicks, and heres what happens:

$u=e^{x^6}$
$du=6x^5e^{x^6}dx$

As a general rule for the future, what exactly does it mean when my du contains my u? What should I do in these scenarios?

$
\int \frac{du}{6} = \frac{1}{6}\int du
$

Would this be the proper way to proceed?
When all I have is my du in my expression, does that always mean to stop "integrating" and to just write it as-is?

Thanks!
u= x^6 not e^{x^6}