∫ [∞,-∞] (x/(x^2+3)^5/2)dx
Evaluate the improper integral and identify wether it converges or diverges.
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
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.
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 ...
To evaluate the improper integral ∫ [∞,-∞] (x/(x^2+3)^(5/2))dx, we can use the technique of symmetry.
Step 1: Rewrite the integral using the property that the integral of an odd function over a symmetric interval is zero. In this case, the function inside the integral is x/(x^2+3)^(5/2), and it is an odd function because it satisfies f(-x) = -f(x).
So, ∫ [∞,-∞] (x/(x^2+3)^(5/2))dx = 0.
Step 2: Determine if the integral converges or diverges. Since the integral evaluates to zero, we can say that the integral converges.
Therefore, the improper integral ∫ [∞,-∞] (x/(x^2+3)^(5/2))dx converges and its value is 0.