# Thread: Didn't know to do this on test

1. ## Didn't know to do this on test

I didn't know how to do this problem on the test I took today...could someone work this out?

Differentiate:

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

2. Originally Posted by c_323_h
I didn't know how to do this problem on the test I took today...could someone work this out?

Differentiate:

$\int \frac{1+x^2}{x^3}dx$
$\frac{1+x^2}{x^3}=\frac{1}{x^3}+\frac{x^2}{x^3}=x^ {-3}+x^{-1}$
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Now I am confused because it say differentiate after integrate. If you really mean that, then it is,
$\frac{1+x^2}{x^3}$ because the derivative of an integral and the integral of the derivative is the same.

3. Originally Posted by ThePerfectHacker
$\frac{1+x^2}{x^3}=\frac{1}{x^3}+\frac{x^2}{x^3}=x^ {-3}+x^{-1}$
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Now I am confused because it say differentiate after integrate. If you really mean that, then it is,
$\frac{1+x^2}{x^3}$ because the derivative of an integral and the integral of the derivative is the same.
yes that's what i thought too! but i haven't seen a "trick" question on a test in years so i thought you could really differentiate it. but shouldn't the answer have a constant added at the end? is is the same with a definite integral?

4. Originally Posted by c_323_h
yes that's what i thought too! but i haven't seen a "trick" question on a test in years so i thought you could really differentiate it. but shouldn't the answer have a constant added at the end? is is the same with a definite integral?
First off, I don't think there are really "trick" questions on tests. (After doing some teaching myself) I suspect that there are teachers that make mistakes and won't fess up to it. I've had some teachers obviously do that, anyway.

Anyway, to the question. When you do an indefinite integral you need to introduce a constant. When you differentiate this the constant goes away and you are left with just the original integrand. When you do a definite integral the result is a number. So when you differentiate this you get zero.

So the answer to your question, as THP says, will be the original integrand.

-Dan

5. Oops looks like I read the directions wrong. It said to evalute the indefinite integral. Well this is the answer...teacher went over the test today:

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

$\frac{x^{-2}}{-2} + ln|x| + C$