Question

the integral of (x/(4x^2+9)^2)

Answer 1

Let's write it like this, it might help us to identify what's going on.
\[\large \int\limits \frac{1}{(\color{orangered}{4x^2+9})^2}(\color{royalblue}{x\;dx})\]

Let \(\large \color{orangered}{u=4x^2+9}\)

Taking the derivative of our substitution gives us,

\(\large du=8x\;dx\)

Understand what's going on so far?

Answer 2

yes

Answer 3

So if we take our du equation and divide each side by 8, we get,\[\large \color{royalblue}{\frac{1}{8}du=x\;dx}\]

Do you see how these pieces will substitute in?
I tried to color-code them to make it easier to see.

Answer 4

yes makes sense

Answer 5

\[\large \int\limits\limits \frac{1}{(\color{orangered}{4x^2+9})^2}(\color{royalblue}{x\;dx}) \qquad = \qquad \int\limits\limits \frac{1}{(\color{orangered}{u})^2}\left(\color{royalblue}{\frac{1}{8}du}\right)\]

Do you understand how to solve it from here?

Answer 6

\[1/8\int\limits_{?}^{?}(1/(u^2)\] what do i do after this

Answer 7

We'll want to recall a rule of exponents:
\(\large x^{-n}=\dfrac{1}{x^n}\)

This works in reverse which is how we'll want to use it.

\[\large \frac{1}{8}\int\limits \frac{1}{u^2}\;du \qquad=\qquad \frac{1}{8}\int\limits u^{-2}\;du\]

And from here we want to apply the `Power Rule for Integration`.
Remember how to do that?

Answer 8

yes thank you so much for your help

Answer 9

what answer did you come out with?

Answer 10

Ummm let's see.\[\large = \frac{1}{8}\left(\frac{u^{-2+1}}{-2+1}\right)+C\]

Which gives usssssss,\[\large -\frac{1}{8}u^{-1}+C \qquad = \qquad -\frac{1}{8u}+C\]From here you need to remember to undo your substitution.
We want our final answer in terms of x, not u.

Answer 11

thank you again. this really helped.

Answer 12

cool, no problem c: