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**An integral with a fraction in it--need help finding the antiderivative**

## Homework Statement

I have the integral [tex]\int(\frac{20}{1+x^2} - 2)dx[/tex] where x = ± 3. To solve, I need to find the antiderivative of [tex]\frac{20}{1+x^2} - 2[/tex]

Now, the worksheet actually gives the answer to the problem: The antiderivative is 37.961 or 37.962. However, I'm having trouble actually reaching that answer on my own; I have trouble with the antiderivatives of fractions.

## Homework Equations

Lessee... I know that the antiderivative of 1/x is ln x. When it gets any more complex than that I get confused.

## The Attempt at a Solution

First, I took [tex]\int(\frac{20}{1+x^2} - 2)dx[/tex] and turned it into [tex]\int(20(1+x^2)^-^1 - 2)dx[/tex]

Then, I tried to take the antiderivative. I thought it would be: [tex]20ln(1+x^2)-2x.[/tex]

Then I plugged in the x values, 3 and -3. With positive 3 plugged in, it came out to 40.052, with a negative 3 it came out to 52.052. I don't even have to finish it off with the subtraction to know that my final answer isn't right and that I missed a step (or two) somewhere. Only question is: What step did I miss? I have a feeling I was supposed to do something with the 20... And I think that there might have been more to taking the antiderivative of (1+x^2)^-1 then just turning it into a natural logarithm. Would someone please help me figure out what I forgot?