Substitution Indefinite Integral Problem

**Jim Marnell** · April 11th 2009, 02:09 PM

Not sure what to do after i get down to the step i'm at, if its all correct so far.

$\int (\frac{x^2+2}{x^3+6x+3})dx$

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

$du=3x^2+6 dx$

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

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

Not sure about or after this:

$\int (\frac{\frac{1}{3}du}{u})$

**nzmathman** · April 11th 2009, 02:23 PM

You don't need substitution...

$\int \frac{x^2+2}{x^3+6x+3}~dx = \frac{1}{3} \int \frac{3x^2+6}{x^3+6x+3}~dx$

Now remember the rule $\int \frac{f'(x)}{f{x}}~dx = \ln{|f(x)|}$ ?

But the last integral you mentioned with "u" will give you the correct answer once you integrate and substitute back. It uses the same rule I mentioned above.

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