How do you evaluate the integral #int x^2sqrt(x-3)#?
1 Answer
I got:
#2/35(x-3)^"3/2"(5x^2 + 12x + 24) + C#
We don't really like radicals, so let's try letting
#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
#= 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)#
