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 $\displaystyle \int\frac{9x^2-2x}{6x^3-2x^2}\,dx=\frac{1}{2}\int\frac{9x^2-2x}{3x^3-x^2}\,dx$
So if $\displaystyle u=3x^3-x^2$, $\displaystyle \,du=9x^2-2x\,dx$
So, $\displaystyle \frac{1}{2}\int\frac{9x^2-2x}{3x^3-x^2}\,dx\xrightarrow{u=3x^3-x^2}{}\frac{1}{2}\int\frac{\,du}{u}$.
Even though this result and the back of the book's result are different, they're actually the same! (can you figure out why?)
Note that if you let $\displaystyle u = 6x^3 - 2x^2$, then $\displaystyle \frac{du}{dx} = 18x^2 - 4x$.
You can rearrange the integrand so that
$\displaystyle \int{\frac{9x^2 - 2x}{6x^3 - 2x^2}\,dx} = \frac{1}{2}\int{\frac{1}{6x^3 - 2x^2}\cdot (18x^2 - 4x)\,dx}$
$\displaystyle = \frac{1}{2}\int{\frac{1}{u}\,\frac{du}{dx}\,dx}$
$\displaystyle = \frac{1}{2}\int{\frac{1}{u}\,du}$
$\displaystyle = \frac{1}{2}\ln{|u|} + C$
$\displaystyle = \frac{1}{2}\ln{|6x^3 - 2x^2|} + C$.
The book is correct.