Integrate x^2(((x^2)+9)^1/2)
I have spent an two hours on this problem I'm still no further. I've tried by parts and by substitution.
If your function has $\displaystyle \begin{align*} a^2 +x^2 \end{align*}$ in it, then a useful substitution is $\displaystyle \begin{align*} x = a\tan{ \left( \theta \right) } \end{align*}$, because it will then simplify to $\displaystyle \begin{align*} a^2 + x^2 = a^2 + \left[ a\tan{ \left( \theta \right) } \right] ^2 = a^2 + a^2 \tan^2{ \left( \theta \right) } = a^2 \left[ 1 + \tan^2{ \left( \theta \right) } \right] = a^2\sec^2{ \left( \theta \right) } \end{align*}$. Why is this useful? Well notice that if $\displaystyle \begin{align*} x = a\tan{ \left( \theta \right) } \end{align*}$ then $\displaystyle \begin{align*} \mathrm{d}x = a\sec^2{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$. So this extra $\displaystyle \begin{align*} \sec^2{ \left( \theta \right) } \end{align*}$ in the differential will either cancel or combine with the one from the substitution, giving you an EASY function to integrate. Now try it!
Yes it's substitution, in this case, trigonometric substitution. The problem is that you have written your integral incorrectly. Every integral needs a differential at the end of it. So your integral is actually $\displaystyle \begin{align*} \int{ x^2 \, \sqrt{ x^2 + 9 } \, \mathrm{d}x } \end{align*}$. You need to replace $\displaystyle \begin{align*} x \end{align*}$ with $\displaystyle \begin{align*} 3\tan{ \left( \theta \right) } \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}x \end{align*}$ with $\displaystyle \begin{align*} 3\sec^2{ \left( \theta \right) } \, \mathrm{d} \theta \end{align*}$. When you have done this, try to simplify the integrand.
Well, I would think you could answer that! When you tried to use that substitution, what went wrong?
Your original integral was $\displaystyle \int x^2(x^2+ 9)^{1/2}dx$. If you make the substitution $\displaystyle u= x^2+ 9$ then du= 2xdx, xdx= du/2. But now, you will need one "x" to use with the differential so that you have $\displaystyle \int x (x^2+ 9)^{1/2} (xdx)= \frac{1}{2}\int x u^{1/2}du$ which still has an "x". You will have to replace that with $\displaystyle x= (u- 9)^{1/2}$ so that once you have converted entirely to the variable u, you have $\displaystyle \int (u- 9)^{1/2}u^{1/2} du$. That's not any easier to integrate than what you started with.
This is one of those problems you can't really get hints for, or possibly even figure out by yourself. As such Prove It has given you the correct substitution to use in post #6. You have a substitution for x in terms of a new variable $\displaystyle \theta$ and what dx is. Using that substitution gives
$\displaystyle \int x^2 \sqrt{x^2 + 9} ~ dx = \int \left ( 3~tan( \theta) \right )^2 \sqrt{ \left ( 3~tan( \theta ) \right ) ^2 + 9}~\left ( 3 ~ sec^2( \theta ) \right )~d \theta $
Now simplify the RHS and see what cancellations/simplifications you can do.
-Dan