Anonymous

# evaluate the indefinite integral of x^2sqt(6-x^3)dx?

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- Anonymous1 decade agoBest Answer
∫ x² √(6 - x^3) dx

let u = 6 - x^3, then du = -3x²dx <==> x²dx = (-1/3)du

substitute

∫√u * (-1/3)du

(-1/3) ∫u^(1/2) du

(-1/3) (2/3) u^(3/2) + C

(-2/9) (6 - x^3)^(3/2) + C <== answer

- 1 decade ago
let u=6-x^3 then you get u'=-3x^2

you already have x^2 in the equation so you just need neg. 3, so multiply neg.3 in the front of the integral sign and multiply the whole thing by -1/3 so that they make one (so they cancel out) then you have integral of u^1/2= 2/3u^(3/2)=2/3(6-x^3)^(3/2) and that's the answer.

- J DLv 51 decade ago
∫x^2sqt(6-x^3)dx

let u = 6-x^3; du = -3x^2 dx

-1/3 ∫sqt(u) du

= -1/3 (2/3)u^(3/2) + C

= -2/9(6-x^3)^(3/2) + C

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- Anonymous1 decade ago
Try

-(2/3) (6 - x^3)^(3/2)

Source(s): Longtime college math teacher

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