A tricky integral

Appa

1. The problem statement, all variables and given/known data
I should find this integral:
[tex]\int[/tex]^b x(1/[tex]\Pi[/tex])(1/(1+x²)dx
^-b

2. Relevant equations

[tex]\int[/tex]1/(1+x²)dx = arctan(x)

3. The attempt at a solution
The Only thing I've succeeded in doing is to take the 1/[tex]\Pi[/tex] and put it in front of the integral like this:
(1/[tex]\Pi[/tex])[tex]\int[/tex]^b (x/(1+x²)dx
^-b
And I know that the integral of 1/(1+x²) equals arctan(x) but how could that help me? Ive tried to use the equation
[tex]\int[/tex]f(x)g(x)dx = F(x)g(x) - [tex]\int[/tex]F(x)g'(x)dx
but I can't compute the integral of arctan(x).
Could someone help me?

Mark44

Is this the integral?
[tex]\frac{1}{\pi}\int_{-b}^b{\frac{x}{x^2 + 1}dx[/tex]

If so, you can use a simple substitution, u = x^2 + 1, and du = 2xdx

Mark44

BTW, you should have posted this in the Calculus & Beyond forum. This problem clearly falls in that area.

NoMoreExams

Seems kind of weird to be doing that integral from -b to b, whether or not b > 0 or < 0 you get into complex numbers.

Gib Z

Perhaps the easiest way is to observe this function is odd, and look at the interval of integration.

HallsofIvy

No, you don't. You are squaring not taking square roots. As Mark44 said, use the substitution u= x²+ 1. Or, even simpler, use Gib Z's suggestion. This is really a very simple integral.

NoMoreExams

Oh duh, good point.

Appa

Yeah, sorry, I got it myself pretty soon after posting this. It feels like the more I study maths, the more I forget..!