# Thread: Another Substitution indefinite integral problem-just needs checked

1. ## Another Substitution indefinite integral problem-just needs checked

Thanks for looking over it and marking any mistakes!
$\displaystyle \int (\frac{x^2+2}{x^3+6x+3})dx$

$\displaystyle u=x^3+6x+3$

$\displaystyle du=3x^2+6(dx)$

$\displaystyle du=3(x^2+2)dx$

$\displaystyle =(x^2+2)dx=\frac{du}{3}$

$\displaystyle \int (\frac{1}{3})(u^-1)$

$\displaystyle \frac{1}{3}\int (u^-1)$

$\displaystyle (\frac{1}{3})(ln{u})+c$

$\displaystyle (\frac{1}{3})(ln{x^3+6x+3})+c$

2. Originally Posted by Jim Marnell
Thanks for looking over it and marking any mistakes!
$\displaystyle \int (\frac{x^2+2}{x^3+6x+3})dx$

$\displaystyle u=x^3+6x+3$

$\displaystyle du=3x^2+6(dx)$

$\displaystyle du=3(x^2+2)dx$

$\displaystyle =(x^2+2)dx=\frac{du}{3}$

$\displaystyle \int (\frac{1}{3})(u^-1)$

$\displaystyle \frac{1}{3}\int (u^-1)$

$\displaystyle (\frac{1}{3})(ln{u})+c$

$\displaystyle (\frac{1}{3})(ln{x^3+6x+3})+c$
Good work.