# Thread: some help with these integrals

1. ## some help with these integrals

ok I'm having problems with these if anyone can help...

integral of x^2(1+x^3)^(-1/3)dx

and this other one

integral between limits of 0 and 1 (6x+6)/(x^2+10x+6)

This ends up in form alog3+blog2 I just need to know how to get there..

2. Originally Posted by dankelly07
ok I'm having problems with these if anyone can help...

integral of x^2(1+x^3)^(-1/3)dx
$\displaystyle \int \frac{x^2}{\sqrt[3]{1-x^3}} \ \mathrm{d}x$

Substituting $\displaystyle u=1-x^3$ and rearranging will simplify the integral to:

$\displaystyle -\frac13 \int \frac{1}{\sqrt[3]u} \ \mathrm{d}u = - \frac13 \int u^{-\frac13} \ \mathrm{d}u$

Solve this integral and substitute $\displaystyle u=1-x^3$ in the final answer.

3. Originally Posted by dankelly07
[snip]
integral between limits of 0 and 1 (6x+6)/(x^2+10x+6)

This ends up in form alog3+blog2 I just need to know how to get there..
Partial fraction decomposition is an obvious option.

4. Originally Posted by dankelly07
ok I'm having problems with these if anyone can help...

integral of x^2(1+x^3)^(-1/3)dx

and this other one

integral between limits of 0 and 1 (6x+6)/(x^2+10x+6)

This ends up in form alog3+blog2 I just need to know how to get there..
Just filling in a couple of steps...

Find $\displaystyle \int \frac{x^2}{\sqrt[3]{1-x^3}} \ \mathrm{d}x$

Let $\displaystyle u=1-x^3$ so $\displaystyle \frac{du}{dx}=-3x^2$

Rearrange your original integral so that you have $\displaystyle \frac{du}{dx}$ as a multiple, in this case by multiplying by a cleverly disguised 1, namely $\displaystyle \frac{-3}{-3}$

Now

$\displaystyle \int \frac{x^2}{\sqrt[3]{1-x^3}} \ \mathrm{d}x = \int \frac{-3x^2}{-3\sqrt[3]{1-x^3}} \ \mathrm{d}x = \frac{-1}{3}\int \frac{1}{\sqrt[3]{u}} \frac{du}{dx}\ dx = \frac{-1}{3}\int \frac{1}{\sqrt[3]{u}}\ du$