Thread: Did I solve this Integral Correctly?

X^2 / third root X+3

u = X+3
du = dx
x= u-3

(u-3)^2 + (u)^-1/3

(u-3)^3 / 3 + (3/2) (u)^2/3

1/3(x+3)^3 + 3/2(x+3)^2/3 + c

Originally Posted by gammaman

X^2 / third root X+3

u = X+3
du = dx
x= u-3

(u-3)^2 + (u)^-1/3

(u-3)^3 / 3 + (3/2) (u)^2/3

1/3(x+3)^3 + 3/2(x+3)^2/3 + c

Your substitution isn't quite right. First, yes, if we let
u = X+3
du = dx

But the next part is highly flawed. Though true that
x = u-3

We're interested in what x^2 is, because that's the numerator. You need to square both sides. Also, when trying to turn this into a product, you need to multiply the numerator by the inverse of the denominator, not add it.

=You need to square both sides. Also, when trying to turn this into a product, you need to multiply the numerator by the inverse of the denominator, not add it.

Can you show me?