# Improper Integration

1. Sep 2, 2009

### Jonathan G

1. The problem statement, all variables and given/known data
$$\int^{\infty}_{0}\frac{x}{(x^{2}+2)^{2}}dx$$

2. Relevant equations
I am well aware how it is to be done but when I take a stab at it, I just can't seem to get the correct solution. I think I might be missing a step somewhere or simply starting off incorrect.

3. The attempt at a solution
$$\int ^{\infty}_{0}\frac{x}{(x^{4}+4x+4)}dx$$

Then I separate into 3 different integrals:

$$\int ^{\infty}_{0}\frac{1}{x^{3}}dx$$ + $$\int ^{\infty}_{0}\frac{1}{(4x)}dx$$ + $$\int ^{\infty}_{0}\frac{x}{4}dx$$

and from there I try solving it the rest of the way but I just can't seem to get a solution that I am satisfied with. The first time I got that it diverges, second time i got divided by zero so I'm not sure which 1 to go with if any.

2. Sep 2, 2009

### rock.freak667

1/(a+b) ≠ 1/a + 1/b

try putting u=x2+2 into the integral.

3. Sep 2, 2009

### Jonathan G

You can't separate the denominator? So I guess it is the numerators you can separate. OK, I'll try this problem again but without expanding the bottom.

4. Sep 2, 2009

### Jonathan G

OK i got it. It came out to converging to -1/4

Thanks!

5. Sep 2, 2009

### Jonathan G

When you have e^(-2*-infinity) it comes out to e^(infinity) hence infinity?

6. Sep 2, 2009

### rock.freak667

$$\lim_{x \rightarrow - \infty} e^{ax} =0$$

EDIT: I corrected it, it is $x \rightarrow \infty$

Last edited: Sep 2, 2009
7. Sep 2, 2009

### Jonathan G

What?!? I thought it was if it was as x-> negative infinity =zero : not when x->positive infinity.

Last edited: Sep 2, 2009
8. Sep 2, 2009

### Staff: Mentor

You're missing an exponent on one of the terms in the denominator. It should be this:
$$\int ^{\infty}_{0}\frac{x}{(x^{4}+4x^2+4)}dx$$

Actually, you didn't do yourself much good by multiplying it out. You could have directly used the substitution that rock.freak667 suggested.
No, no, no! You really should go back and review how fractions and rational expressions add.