Integrals

**xwrathbringerx** · September 25th 2009, 10:14 PM

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!!!

**Chris L T521** · September 25th 2009, 10:20 PM

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?)

**Prove It** · September 25th 2009, 10:22 PM

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.

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