# A tricky integral

by Appa
Tags: arctan(x), integral
 P: 16 1. The problem statement, all variables and given/known data I should find this integral: $$\int$$b x(1/$$\Pi$$)(1/(1+x2)dx -b 2. Relevant equations $$\int$$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/$$\Pi$$ and put it in front of the integral like this: (1/$$\Pi$$)$$\int$$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 $$\int$$f(x)g(x)dx = F(x)g(x) - $$\int$$F(x)g'(x)dx but I can't compute the integral of arctan(x). Could someone help me?
 Mentor P: 19,710 Is this the integral? $$\frac{1}{\pi}\int_{-b}^b{\frac{x}{x^2 + 1}dx$$ If so, you can use a simple substitution, u = x^2 + 1, and du = 2xdx
 Mentor P: 19,710 BTW, you should have posted this in the Calculus & Beyond forum. This problem clearly falls in that area.
P: 626

## A tricky integral

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.
 HW Helper P: 3,353 Perhaps the easiest way is to observe this function is odd, and look at the interval of integration.
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