I didn't know how to do this problem on the test I took today...could someone work this out?
Differentiate:
$\displaystyle \int \frac{1+x^2}{x^3}dx$
$\displaystyle \frac{1+x^2}{x^3}=\frac{1}{x^3}+\frac{x^2}{x^3}=x^ {-3}+x^{-1}$Originally Posted by c_323_h
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Now I am confused because it say differentiate after integrate. If you really mean that, then it is,
$\displaystyle \frac{1+x^2}{x^3}$ 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?Originally Posted by ThePerfectHacker
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.Originally Posted by c_323_h
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
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:
$\displaystyle \int\frac{1+x^2}{x^3}dx =$
$\displaystyle \int\frac{1}{x^3}+\frac{x^2}{x^3}dx = $
$\displaystyle \int x^{-3} dx+\int\frac{1}{x}dx =$
$\displaystyle \frac{x^{-2}}{-2} + ln|x| + C$