## How Can An Integration Result In Different Expressions?

\[(x-1)^2\]

,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

\[c_1=c_2- \frac{1}{3}\]

.