What is the integral of (x^5)*(x^2 - 9)^(3/2)

[From: ] [author: ] [Date: 12-08-18] [Hit: ]

So, x^2 = u + 9 and du = 2x dx.Hence,Expanding,= (1/9) (x^2 - 9)^(9/2) + (18/7) (x^2 - 9)^(7/2) + (81/5) (x^2 - 9)^(5/2) + C.I hope this helps!......

I know you are suppose to use u substitution

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This one is a bit subtle.

Let u = x^2 - 9.
So, x^2 = u + 9 and du = 2x dx.

Hence, ∫ x^5 (x^2 - 9)^(3/2) dx
= ∫ (x^2)^2 (x^2 - 9)^(3/2) * x dx
= ∫ (u + 9)^2 * u^(3/2) * du/2

Expanding, we obtain
(1/2) ∫ (u^2 + 18u + 81) * u^(3/2) du
= (1/2) ∫ (u^(7/2) + 18u^(5/2) + 81u^(3/2)) du
= (1/2) [(2/9) u^(9/2) + 18 * (2/7) u^(7/2) + 81 * (2/5) u^(5/2)] + C
= (1/9) (x^2 - 9)^(9/2) + (18/7) (x^2 - 9)^(7/2) + (81/5) (x^2 - 9)^(5/2) + C.

I hope this helps!

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