∫ [∞,-∞] (x/(x^2+3)^5/2)dx

Question

Evaluate the improper integral and identify wether it converges or diverges.

Answer 1

That pesky √(x^2+3) is a sure sign you can use a trig substitution, such as
x^2 = 3tan^2θ
x^2+3 = 3sec^2θ
2x dx = 6 sec^2θ tanθ dθ
Now you have
1/2 ∫ (x^2+3)^(-5/2) * 2xdx
= 1/2 (√3 secθ)^-5 * 6 secθ * secθ tanθ dθ
= 1/(9√3) ∫ (secθ)^-4 d(secθ)
= -1/3 (3secθ)^-3
= -1/3 (x^2+3)^(-3/2)

Since f(x) is an odd function, the integral will be 0.

Answer 2

or ...

since I see the derivative of the base (x^2 + 3) hanging around as a multiple, I can use the "just observation method"

∫ x(x^2 + 3)^(5/2) dx
= x (x^2 + 3)^(7/2) / (2x * 7/2)
= (1/7)(x^2 + 3)^(7/2) + c

or, using the simple substitution,
let u = x^2 +3
du = 2x dx
x dx = du / 2

∫ x(x^2 + 3)^(5/2) dx
= ∫ u^(5/2) du/2
= (1/2)(7/2) u^(7/2)
= 1/7(x^2 + 3)^(7/2) + c

Answer 3

well, duh. All that cool trig work for nothing.
As the Rock Man said,
"You see what you wanta see;
You hear what you wanta hear."

One thing I always tell my students is to look first for the easy substitution ...