# 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

Staff Emeritus
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..!