$\displaystyle \int 2(x+1)dx$

Method 1

$\displaystyle u=x+1$

$\displaystyle du=dx$

$\displaystyle \int 2udu$

$\displaystyle u^2+c$

$\displaystyle (x+1)^2+c$

Method 2

$\displaystyle \int (2x+2) dx$

$\displaystyle x^2+2x+c$

???

- Apr 22nd 2010, 05:54 PMchengbin2 integration method giving different results? Where did I go wrong?
$\displaystyle \int 2(x+1)dx$

Method 1

$\displaystyle u=x+1$

$\displaystyle du=dx$

$\displaystyle \int 2udu$

$\displaystyle u^2+c$

$\displaystyle (x+1)^2+c$

Method 2

$\displaystyle \int (2x+2) dx$

$\displaystyle x^2+2x+c$

??? - Apr 22nd 2010, 05:59 PMskeeter