Question: How do you find the integral of ((1+x^2)^1/2)x^5 dx ?
Use integration by parts.
I = int (1+x^2)^(1/2).x^5 dx = (1/3).(1+x^2)^(3/2).x^4 - (4/3).int (1+x^2)^(3/2).x^3 dx, with
I2 = int (1+x^2)^(3/2).x^3 dx = (1/5).(1+x^2)^(5/2).x^2 - (2/5).int (1+x^2)^(5/3).x dx, with
I3 = int (1+x^2)^(5/3).x dx = (1/7).(1+x^2)^(7/2)
So, I = (1/3).(1+x^2)^(3/2).x^4 - (4/3).{(1/5).(1+x^2)^(5/2).x^2 - (2/5).[(1/7).(1+x^2)^(7/2)]}
I = (1/3).(1+x^2)^(3/2).x^4 - (4/15).(1+x^2)^(5/2).x^2 + (8/105 ).(1+x^2)^(7/2)
I = √(1+x^2)/105.{35(1+x^2).x^4 - 28(1+x^2)^2.x^2 + 8(1+x^2)^3}
Answer: I = (1/105).(1+x^2)^(3/2).{8 - 12*x^2 + 15*x^4)