Integral through $u$-substitution v. multiplying out I have the integral:  
$ \int (x^2)(x^3-1)^2 \, dx $  Through $u$-substitution, I can write this is equal to $ \frac {1}{3} \int (3x^2)(x^3-1)^2 dx $  which equals $ \frac{1}{3}\cdot\frac{(x^3-1)^3}{3}$. 
However, if rather than using $u$-substitution, I multiply within $ \int (x^2)(x^3-1)^2 \, dx $ from the beginning, I don't get the same answer. For example: $ \int (x^2)(x^3-1)^2 \, dx $ = $ \int (x^2)(x^6 -2x^3 + 1) \, dx = \int (x^8 -2x^5 + x^2) \, dx$. The anti derivative of this becomes $ \frac{x^9}{9} - \frac{2x^6}{6} + \frac{x^3}{3} $.   
However, these anti-derivatives are not equal. Through u-substitution, when $x=1$, the anti derivative evaluates to zero. In the second route, when I plug in $1$, I get $1/9$.
Where did I go wrong? From what I see, the $u$-substitution version is correct. 

Ahhh, the question is what is the function at the point (1,0). And the constant will change if I use the first equation v. the second. So the first equation will have a constant of 0, which has a solution function of $ \frac {1}{3} * \frac {(x^3-1)^3}{3}$ with its constant = 0. 
And the second way  $ \frac{x^9}{9} - \frac{2x^6}{6} + \frac{x^3}{3} - \frac{1}{9} $ where 1/9 is my constant. I assume now the two functions are equal. 
