# A tricky integral

by Appa
Tags: arctan(x), integral
 Share this thread:
 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: 21,409 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: 21,409 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: 3,348 Perhaps the easiest way is to observe this function is odd, and look at the interval of integration.
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,688
 Quote by 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.
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.
 P: 626 Oh duh, good point.
 P: 16 Yeah, sorry, I got it myself pretty soon after posting this. It feels like the more I study maths, the more I forget..!

 Related Discussions Calculus 9 Calculus & Beyond Homework 3 Introductory Physics Homework 18 Calculus 2 Calculus 7