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

1 Answer
Truong-Son N.
Feb 11, 2017

I got:

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

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)#

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