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

Differentiate:

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

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- Jul 5th 2006, 07:28 PMc_323_hDidn'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:

$\displaystyle \int \frac{1+x^2}{x^3}dx$ - Jul 5th 2006, 07:36 PMThePerfectHackerQuote:

Originally Posted by**c_323_h**

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Now I am confused :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. - Jul 6th 2006, 03:10 AMc_323_hQuote:

Originally Posted by**ThePerfectHacker**

- Jul 6th 2006, 06:05 AMtopsquarkQuote:

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 - Jul 6th 2006, 06:51 AMc_323_h
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 = $

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

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