Another Substitution indefinite integral problem-just needs checked

**Jim Marnell** · April 14th 2009, 09:48 AM

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

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

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

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

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

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

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

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

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

**Chop Suey** · April 14th 2009, 10:06 AM

Originally Posted by Jim Marnell

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

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

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

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

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

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

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

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

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

Good work.

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