An integral with a fraction in it- finding the antiderivative

[BZ908]

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

Homework Statement

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

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 \int(\frac{20}{1+x^2} - 2)dx and turned it into \int(20(1+x^2)^-^1 - 2)dx

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

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?

 

[Vid]

Try a trig sub or look up derivatives of inverse trig functions.

 

[Defennder]

If you're using a trigo substitution, you may want to use this:
tan^2\theta + 1 = sec^2\theta

That should help you.

 

[Feldoh]

Don't be a victim of universal logarithmic differentiation!

\int\frac{20}{1+x^2}dx doesn't equal 20*ln|1+x^2|+c

 

[rootX]

If you haven't touched trig int before:

let x = tanu
dx = tanu.secu

substitute these values for x and dx

 

[BrendanH]

Oh hush! We all know it's 20arctanx - 2x! Cut the kid some slack!

The reason the natural logarithm didn't work was because you need the derivative of the denominator in the numerator. The derivative of 1+x^2 is 2x, and as you can see, it ain't there.

 

[rootX]

Oh hush! We all know it's 20arctanx - 2x! Cut the kid some slack!

The reason the natural logarithm didn't work was because you need the derivative of the denominator in the numerator. The derivative of 1+x^2 is 2x, and as you can see, it ain't there.

I didn't know it three-four years ago :shy:...