Evaluate the Indefinite Integral

• Apr 9th 2009, 09:49 PM
Jim Marnell
Evaluate 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 PM
mollymcf2009
Quote:

Originally Posted by Jim Marnell
$\displaystyle \int{\frac{\sqrt{x}-x^2+4}{x}}dx$

Lost on what to do.
Thanks for the help!

SImplify it first.

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

Then integrate term by term. Can you try it from here?
• Apr 9th 2009, 10:05 PM
Jim Marnell
thanks! ill try it from here and post what i get.
• Apr 10th 2009, 05:24 AM
Jim 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 AM
mr fantastic
Quote:

Originally Posted by Jim Marnell
$\displaystyle \int{\frac{\sqrt{x}-x^2+4}{x}}dx$
$\displaystyle \int (\frac{1}{\sqrt{x}} - x + \frac{4}{x}) dx$
$\displaystyle \int (-x^{1/2}-x+\frac{4}{x}) dx$ Mr F says: No. You have to use your basic index laws. It's $\displaystyle {\color{red}\int \! x^{-1/2} - x + 4 x^{-1} \, dx}$.

Would that be the correct step, i'm lost after that. Not sure how to get the antiderivatives to evaluate

You're expected to know that $\displaystyle \int \! a x^{n} \, dx = \frac{a}{n+1} x^{n+1} + C$, provided $\displaystyle n \neq -1$.
• Apr 10th 2009, 06:00 AM
Jim 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 PM
mr fantastic
Quote:

Originally Posted by Jim 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

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 AM
foodmmm
what is the final answer to this?
• Apr 14th 2009, 06:19 AM
mr fantastic
Quote:

Originally Posted by foodmmm
what is the final answer to this?

Note: $\displaystyle \frac{- x^{3/2}}{3/2} = - \frac{2}{3} x^{2/3}$.