How do I find the integral of x(x^2 -4)^(7/2)?

Answer 1

I = ∫ x (x² - 4)^(7/2) dx
Let u = x² - 4
du = 2x dx
du / 2 = x dx
I = (1/2) ∫ u^(7/2) du
I = (1/2) [(2/9) u^(9/2)] + C
I = (1/9) (x² - 4)^(9/2) + C

Answer 2

Int x(x^2 -4)^(7/2)

Let u = x^2 -4
Then du = 2x dx

Int x(x^2 -4)^(7/2) dx
=Int 1/2 u^(7/2) du
=1/2 u ^(9/2) / (9/2) +k
=1/9 u ^(9/2) + k
= (x^2 -4)/9 + k

Answer 3

∫x(x^2 -4)^(7/2) dx
= ∫(1/2)(x^2 -4)^(7/2)d(x^2-4), since xdx = (1/2)d(x^2-4)
= (1/9)(x^2-4)^(9/2) + c