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