## How Can An Integration Result In Different Expressions?

There is more than one way to integrate
$(x-1)^2$
,
You could expand the brackets and integrate term by term.
$\int x^2-2x+1 dx=\frac{x^3}{3}-x^2+x+c$

or you could use the substitution
$u=x-1$
(then
$du=dx$
) to get
$\int u^2 du=\frac{u^3}{3}+c=\frac{(x-1)^3}{3}+c$

Obviously both expressions are not the same. How can they both be correct?
The arbitrary constants
$c$
are not the same!
If we expand the brackets for the second answer we get
$\frac{(x-1)^3}{3}+c=\frac{x^3-3x^2+3x-1}{3}+c=\frac{x^3}{3}-x^2+ x- \frac{1}{3}+c$
.
Now both expressions are the same except for the constant terms.
In fact we can write
$\int x^2+-2x+1 dx=\frac{x^3}{3}-x^2+x+c_1$

$\int u^2 du=\frac{u^3}{3}+c=\frac{(x-1)^3}{3}+c_2$

Where
$c_1=c_2- \frac{1}{3}$
.