text: # Homework Help: A tricky integral
  
  1. Jan 14, 2009
  
  ### Appa
  
  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?
  
  2. Jan 14, 2009
  
  ### Staff: Mentor
  
  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
  
  3. Jan 14, 2009
  
  ### Staff: Mentor
  
  BTW, you should have posted this in the Calculus & Beyond forum. This problem clearly falls in that area.
  
  4. Jan 14, 2009
  
  ### 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.
  
  5. Jan 14, 2009
  
  ### Gib Z
  
  Perhaps the easiest way is to observe this function is odd, and look at the interval of integration.
  
  6. Jan 14, 2009
  
  ### 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.
  
  7. Jan 14, 2009
  
  ### NoMoreExams
  
  Oh duh, good point.
  
  8. Jan 14, 2009
  
  ### Appa
  
  Yeah, sorry, I got it myself pretty soon after posting this. It feels like the more I study maths, the more I forget..!