integrate ((x^2-4)^2))^(1/2)
Would the 2 and the 1/2 cancel
yes. so you are left with just (x^2-4)
as usual, (√x)^2 = √(x^2) = x
then how would I solve because the derivative of x^2-4 =2x and in the original there is only a dx.
the original should be ((x^2-4)^2)^(1/2) dx
(z^2)^(1/2) = z
you have
(x^2 - 4) dx
x^3/3 - 4 x + C
To integrate the function ((x^2-4)^2)^(1/2), we need to simplify it first. The expression ((x^2-4)^2)^(1/2) represents the square root of the square of (x^2-4).
To simplify, we can square ((x^2-4)^2)^(1/2):
((x^2-4)^2)^(1/2) = (x^2-4)^(2*(1/2)) = (x^2-4)^1 = (x^2-4).
Therefore, the given function can be simplified to (x^2-4).
Now, we can integrate (x^2-4) using standard integration techniques. The integral of (x^2-4) can be found by applying the power rule of integration. The power rule states that for any function of the form x^n, the integral is (x^(n+1))/(n+1), where n is not equal to -1.
So, integrating (x^2-4):
∫ (x^2-4) dx = ∫ x^2 dx - ∫ 4 dx
Now, let's find the integrals of each term separately:
∫ x^2 dx = (1/3)x^3 + C1 (where C1 is the constant of integration)
∫ 4 dx = 4x + C2 (where C2 is the constant of integration)
Putting it all together:
∫ (x^2-4) dx = (1/3)x^3 - 4x + C (where C is the constant of integration)
Therefore, the integral of ((x^2-4)^2)^(1/2) is equal to (1/3)x^3 - 4x + C.