\[(x-1)^2\]
,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
\[c_1=c_2- \frac{1}{3}\]
.