Re: Integration by substitution
Ouch, that is a tough one!
But I didn't get around to solving "Student"s integral. Which I will attempt now ...
∫ x^2/(1 - 2x^3) dx
Then: du = -6x^2 dx (ie we differentiated it)
So: x^2 dx = (-1/6) du
Substituting 1-2x^3=u and x^2 dx = (-1/6) du we get:
∫ x^2/(1 - 2x^3) dx = ∫ 1/u (-1/6) du
Then it is fairly easy to solve ∫ 1/u (-1/6) du:
∫ 1/u (-1/6) du = (-1/6) ∫ 1/u du = (-1/6) ln u
Now put back u=1-2x^3:
(-1/6) ln( 1-2x^3 )
DONE! (I hope!)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman