1. ## Antiderivative Question

The directions state: Find the indefinite integral and check the result by differentiation.

Here is the equation. (Please open the link below. I drew the equation on a paint program. I did not know how to post otherwise. Sorry.)
http://i34.tinypic.com/24qntol.jpg

If someone could please try this problem. When I did this I got an answer that did not check correctly with the original equation.
Thank you!!!

2. $\frac{x^2+2x-3}{x^4} = \frac{x^2}{x^4} + \frac{2x}{x^4} - \frac{3}{x^4} = x^{-2} + 2x^{-3} + 3x^{-4}$

can you integrate the last expression?

3. Split this up into seperate integrals by distributing the denominator of the fraction,

$\int\frac{x^2+2x-3}{x^4}dx=\int\frac{x^2}{x^4}dx+\int\frac{2x}{x^4} dx-\int\frac{3}{x^4}dx$.

Now can you do it?

4. ## Mine

That's what I got too. I ended up getting:
http://i37.tinypic.com/308v5o2.jpg

I went and checked and I couldn't get the -x^-1 to work right. Could some one check me please?

5. $\frac{d}{dx}(-x^{-1}-x^{-2}+x^{-3})=x^{-2}+2x^{-3}-3x^{-4}$

so

$\frac{df}{dx}=\frac{1}{x^2}+\frac{2}{x^3}-\frac{3}{x^4}=\frac{x^2}{x^4}+\frac{2x}{x^4}-\frac{3}{x^4}=\frac{x^2+2x-3}{x^4}$.

You were right.

6. ## Antiderivate

I understand how to rewrite the equation so I can take the antiderivative but I guess what I have a question about now is taking the antiderivative....

7. ## Thanks

Oh I was right?

That makes me feel better. Thank you so much!