Question

How do you evaluate the integral `int x^2sqrt(x-3)`?

Answer 1

I got:

`2/35(x-3)^"3/2"(5x^2 + 12x + 24) + C`

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We don't really like radicals, so let's try letting `u = sqrt(x-3)`. Then:

`du = 1/(2sqrt(x-3))dx => dx = 2udu`
`(u^2 + 3)^2 = x^2`

Therefore, we have:

`int 2u(u^2+3)^2udu`

`= 2int u^2(u^2+3)^2du`

`= 2int u^2(u^4 + 6u^2 + 9)du`

`= 2int u^6 + 6u^4 + 9u^2du`

Now this is straightforward to integrate.

`= 2(u^7/7 + 6/5u^5 + 3u^3)`

`= 2/7 u^7 + 12/5u^5 + 6u^3`

Sub `u = sqrt(x-3)` back in to get:

`= 2/7 (x-3)^"7/2" + 12/5(x-3)^"5/2" + 6(x-3)^"3/2" + C`

This is acceptable, but we could also simplify further.

`= (x-3)^"3/2"[2/7 (x-3)^2 + 12/5(x-3) + 6] + C`

`= (x-3)^"3/2"[2/7 (x^2 - 6x + 9) + 12/5x - 36/5 + 6] + C`

`= (x-3)^"3/2"(2/7x^2 - 12/7x + 18/7 + 12/5x - 36/5 + 30/5) + C`

`= (x-3)^"3/2"(10/35x^2 - 60/35x + 90/35 + 84/75x - 252/35 + 210/35) + C`

`= (x-3)^"3/2"(10/35x^2 + 24/35x + 48/35) + C`

`= color(blue)(2/35(x-3)^"3/2"(5x^2 + 12x + 24) + C)`