Thread: Integrals

1. Integrals

Hi if you integrate (9x^2-2x)/(6x^3-2x^2), what's the answer? The textbook says 1/2 * log(6x^3-2x^2) + c...is that correct? Because I keep on getting 3/4 * log (6x^2 - 2x) + c

PLEASE HELP!!!

2. Originally Posted by xwrathbringerx
Hi if you integrate (9x^2-2x)/(6x^3-2x^2), what's the answer? The textbook says 1/2 * log(6x^3-2x^2) + c...is that correct? Because I keep on getting 3/4 * log (6x^2 - 2x) + c

PLEASE HELP!!!
Note that $\int\frac{9x^2-2x}{6x^3-2x^2}\,dx=\frac{1}{2}\int\frac{9x^2-2x}{3x^3-x^2}\,dx$

So if $u=3x^3-x^2$, $\,du=9x^2-2x\,dx$

So, $\frac{1}{2}\int\frac{9x^2-2x}{3x^3-x^2}\,dx\xrightarrow{u=3x^3-x^2}{}\frac{1}{2}\int\frac{\,du}{u}$.

Even though this result and the back of the book's result are different, they're actually the same! (can you figure out why?)

3. Originally Posted by xwrathbringerx
Hi if you integrate (9x^2-2x)/(6x^3-2x^2), what's the answer? The textbook says 1/2 * log(6x^3-2x^2) + c...is that correct? Because I keep on getting 3/4 * log (6x^2 - 2x) + c

PLEASE HELP!!!
Note that if you let $u = 6x^3 - 2x^2$, then $\frac{du}{dx} = 18x^2 - 4x$.

You can rearrange the integrand so that

$\int{\frac{9x^2 - 2x}{6x^3 - 2x^2}\,dx} = \frac{1}{2}\int{\frac{1}{6x^3 - 2x^2}\cdot (18x^2 - 4x)\,dx}$

$= \frac{1}{2}\int{\frac{1}{u}\,\frac{du}{dx}\,dx}$

$= \frac{1}{2}\int{\frac{1}{u}\,du}$

$= \frac{1}{2}\ln{|u|} + C$

$= \frac{1}{2}\ln{|6x^3 - 2x^2|} + C$.

The book is correct.