$\displaystyle \int{\frac{\sqrt{x}-x^2+4}{x}}dx$

Lost on what to do.

Thanks for the help!

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- Apr 9th 2009, 09:49 PMJim MarnellEvaluate the Indefinite Integral
$\displaystyle \int{\frac{\sqrt{x}-x^2+4}{x}}dx$

Lost on what to do.

Thanks for the help! - Apr 9th 2009, 10:03 PMmollymcf2009
- Apr 9th 2009, 10:05 PMJim Marnell
thanks! ill try it from here and post what i get.

- Apr 10th 2009, 05:24 AMJim Marnell
$\displaystyle \int{\frac{\sqrt{x}-x^2+4}{x}}dx$

$\displaystyle \int (\frac{1}{\sqrt{x}} - x + \frac{4}{x}) dx$

$\displaystyle \int (-x^{1/2}-x+\frac{4}{x}) dx$

Would that be the correct step, i'm lost after that. Not sure how to get the antiderivatives to evaluate - Apr 10th 2009, 05:36 AMmr fantastic
- Apr 10th 2009, 06:00 AMJim Marnell
so would this be the next step:

$\displaystyle \int (\frac{-x^{3/2}}{3/2}-\frac{x^2}{2}+\frac{4}{x})dx$

not sure how to find the antiderivative of 4/x - Apr 10th 2009, 01:34 PMmr fantastic
Since this answer means you have done the integral, the integration operator should not be included. So you have

$\displaystyle \frac{-x^{3/2}}{3/2} - \frac{x^2}{2} + \int \! \frac{4}{x} \, dx$.

The last integral is equal to $\displaystyle 4 \ln |x | $ (again, you're expected to know this.)

Your final answer needs to include a "+ C" (and again, this is something you're meant to know). - Apr 14th 2009, 05:59 AMfoodmmm
what is the final answer to this?

- Apr 14th 2009, 06:19 AMmr fantastic