# Thread: Need help integrating this

1. ## Need help integrating this

Hi can someone help me integrate this:

$\int \frac{1}{4}x^{2}\frac{1}{\sqrt{x^{3}+1}}dx$

I must have forgotten, but do I use u substitution?

2. ## Re: Need help integrating this

Yes, start with $u = x^3, du = 3x^2 dx$
This becomes
$\frac{1}{12} \int \frac{1}{\sqrt{u} +1} du$

To integrate this im not really sure off the top of my head how to do it formally

$= \frac{1}{12}( 2\sqrt{u} -2log(\sqrt{u}+1))$

err had an arithmetic error.

3. ## Re: Need help integrating this

Thank you very much scopur!

I'm really sorry but I just realised that the 1 was meant to be under the square-root symbol. I have made he changes, hopefully that doesn't that affect anything too much?

4. ## Re: Need help integrating this

With practice, this sort of problem is solved by differentiation rather than integration.
Look at the $x^{2}$ on the top line and the $\sqrt{x^{3}+1}$ on the bottom line and realise that they have come from differentiating $\sqrt{x^{3}+1}.$
So, differentiate $\sqrt{x^{3}+1}=(x^{3}+1)^{1/2}$ function of a function fashion to get $(1/2)(x^{3}+1)^{-1/2}.3x^{2}$ and then multiply by a suitable constant to get the required result.
With just a little practice, you just write down the result.

If you prefer the integration by substitution, $u^{2}=x^{3}+1$ works best.

5. ## Re: Need help integrating this

Are you sure you'd let \displaystyle \begin{align*} u^2 = x^3 + 1 \end{align*}? I think \displaystyle \begin{align*} u = x^3 + 1 \implies du = 3x^2 \, dx \end{align*} works easier...

\displaystyle \begin{align*} \int{\frac{1}{4}\, x^2\, \frac{1}{\sqrt{x^3 + 1}}\,dx} &= \frac{1}{12} \int{ 3x^2\, \frac{1}{\sqrt{x^3 + 1}}\, dx } \\ &= \frac{1}{2} \int{\frac{1}{\sqrt{u}}\,du} \\ &= \frac{1}{2} \int{u^{-\frac{1}{2}}\,du} \end{align*}

Go from here...

6. ## Re: Need help integrating this

If $u^{2}=x^{3}+1,$ then $2u\, du = 3x^{2}dx$ so

$\int \frac{1}{4}\frac{x^{2}}{\sqrt{x^{3}+1}}\,dx=\frac{ 1}{12}\int \frac{1}{\sqrt{x^{3}+1}}\,3x^{2}\,dx=\frac{1}{12} \int \frac{1}{u}\,2u\,du = \frac{1}{12} \int 2 \, du,$ etc..