solving integrals?

Question

how do you do this integral?

integration [(x^8)/((4-x^2)^(11/2))dx]

Answer 1

Integral [x^8 / (4-x^2)^11/2] dx …. (1)

Let x = 2 sin z, ---- (2)
so dx = 2 cos z dz,
note also (1 – sin z^2)^11/2 = cos z^11 , (4)^11/2 = 2^11

Integral (2 cos z dz) [(2)^8* sin z^8 / (2^11)(cos z)^11]

Integral [sin z^8 / (2^2)(cos z)^10 dz]

Integral (1/4) [tan z^8 * sec z)^2 dz] ---- (3)

Again let p = tan z, dp = secz^2 . dz put in (3) and

Integral (1/4) [p^8 dp]
= (1/4) * (1/9) p^9 = (1/36) p^9 = (1/36) tan z^9 ------ (4)

from eq (2), x/2 = sin z therefore cos z = (1/2)* sqrt (4-x^2)

or tan z = x / sqrt (4-x^2) put back in eq (4)

Answer = (1/36)* {x / sqrt (4-x^2)}^9

Answer 2

Ouch, I think the only way to do it would be integration by parts several times.

Let u = x^7 and dv = x / ((4-x^2)^(11/2))

Then let u = x^5 and dv = x / ((4-x^2)^(9/2))

etc....