Hi can someone help me integrate this:
$\displaystyle \int \frac{1}{4}x^{2}\frac{1}{\sqrt{x^{3}+1}}dx$
I must have forgotten, but do I use u substitution?
Yes, start with$\displaystyle u = x^3, du = 3x^2 dx $
This becomes
$\displaystyle \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
$\displaystyle = \frac{1}{12}( 2\sqrt{u} -2log(\sqrt{u}+1)) $
err had an arithmetic error.
With practice, this sort of problem is solved by differentiation rather than integration.
Look at the $\displaystyle x^{2}$ on the top line and the $\displaystyle \sqrt{x^{3}+1}$ on the bottom line and realise that they have come from differentiating $\displaystyle \sqrt{x^{3}+1}.$
So, differentiate $\displaystyle \sqrt{x^{3}+1}=(x^{3}+1)^{1/2}$ function of a function fashion to get $\displaystyle (1/2)(x^{3}+1)^{-1/2}.3x^{2}$ 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, $\displaystyle u^{2}=x^{3}+1$ works best.
Are you sure you'd let $\displaystyle \displaystyle \begin{align*} u^2 = x^3 + 1 \end{align*}$? I think $\displaystyle \displaystyle \begin{align*} u = x^3 + 1 \implies du = 3x^2 \, dx \end{align*}$ works easier...
$\displaystyle \displaystyle \begin{align*} \int{\frac{1}{4}\, x^2\, \frac{1}{\sqrt{x^3 + 1}}\,dx} &= \frac{1}{12} \int{ 3x^2\, \frac{1}{\sqrt{x^3 + 1}}\, dx } \\ &= \frac{1}{2} \int{\frac{1}{\sqrt{u}}\,du} \\ &= \frac{1}{2} \int{u^{-\frac{1}{2}}\,du} \end{align*}$
Go from here...
If $\displaystyle u^{2}=x^{3}+1,$ then $\displaystyle 2u\, du = 3x^{2}dx$ so
$\displaystyle \int \frac{1}{4}\frac{x^{2}}{\sqrt{x^{3}+1}}\,dx=\frac{ 1}{12}\int \frac{1}{\sqrt{x^{3}+1}}\,3x^{2}\,dx=\frac{1}{12} \int \frac{1}{u}\,2u\,du = \frac{1}{12} \int 2 \, du,$ etc..