I have been having a trouble with this problem:
X / (4X^2+1)^2 dx<-------This is an integral that I have to solve using substitution
This is what I did.
u=4x^2+1
u'=8x
du/8=dx
So then I went back to original problem and plugged that in and got.
1/8*(1/(1/3)u^3) I got that from taking the anti derivative of x^2
I plugged u back in and got
1/8 * 1/(1/3)(4x^2+1)^3
after i simplify i get:
1/ ((8/3)(64x^6+1))
I would appreciate it if someone could tell me where i messed up. Thanks!
I got that far. I then took the anti derivative of u^-2 and got -U^-1
I plugged u into that equation and got -((4x^2)+1)^(-1+2) = -((4x^2)+1)+c Then I multiply that by 1/8 which gives:
-((4x^2)+1)+c / 8
I must be messing up with the algebra part because I kept getting the same thing you posted.
My teacher gave -1/(32x^2+8) + c as the correct answer.