# Thread: Evaluate the Indefinite Integral

1. ## Evaluate the Indefinite Integral

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

Lost on what to do.
Thanks for the help!

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

Lost on what to do.
Thanks for the help!

SImplify it first.

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

Then integrate term by term. Can you try it from here?

3. thanks! ill try it from here and post what i get.

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

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

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

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 $\int \! a x^{n} \, dx = \frac{a}{n+1} x^{n+1} + C$, provided $n \neq -1$.

6. so would this be the next step:
$\int (\frac{-x^{3/2}}{3/2}-\frac{x^2}{2}+\frac{4}{x})dx$

not sure how to find the antiderivative of 4/x

7. Originally Posted by Jim Marnell
so would this be the next step:
$\int (\frac{-x^{3/2}}{3/2}-\frac{x^2}{2}+\frac{4}{x})dx$

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

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

The last integral is equal to $4 \ln |x |$ (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).

8. what is the final answer to this?

9. Originally Posted by foodmmm
what is the final answer to this?
Note: $\frac{- x^{3/2}}{3/2} = - \frac{2}{3} x^{2/3}$.