Integration of (9x^2-2x)/(6x^3-2x^2) respective to x.
I keep on getting 3/4 * ln(x^2-3) + c
Is that right???? Could someone please show me how to get the right answer if it's wrong?
Use the substitution $\displaystyle u = 3x^3 - x^2$ so that $\displaystyle \frac{du}{dx} = 9x^2 - 2x$.
Therefore
$\displaystyle \int{\frac{9x^2 - 2x}{6x^3 - 2x^2}\,dx} = \frac{1}{2}\int{\frac{1}{u}\,\frac{du}{dx}\,dx}$
$\displaystyle = \frac{1}{2}\int{\frac{1}{u}\,du}$
$\displaystyle = \frac{1}{2}\ln{|u|} + C$
$\displaystyle = \frac{1}{2}\ln{|3x^3 - x^2|} + C$.