Integration by parts help $\int^3_{-1}\frac{1}{2}e^{\sqrt{x+1}}\,dx$ 
Use the substitution $u=\sqrt{x+1}$ to find the exact value of 
  $\displaystyle\int_{-1}^3 \frac{1}{2}e^{\sqrt{x+1}}\,dx.$

I've substituted and found the derivative of u, but I'm not sure what to do when I substitute it back in. Can someone tell me every step I need to do?
 A: The first step is to differentiate the substitution:
$\displaystyle u = \sqrt {x + 1} $
$\displaystyle \frac {du} {dx} = \frac d {dx} [(x + 1)^{\frac 1 2}] $
$\displaystyle \frac {du} {dx} = \frac {1} {2 \sqrt {x + 1}} $
The next step is determining the infinitesimal relationship:
$\displaystyle du = \frac {1} {2 \sqrt {x + 1}} dx $
Note however, that substitution at this point won't help:
$\displaystyle \int _{-1} ^{3} {\frac 1 2 e^{\sqrt{(x + 1)}}} \ dx = \int _{-1} ^{3} {\frac 1 2 e^u} \ dx $
The addition "trick" is the substitution in the infinitesimal relationship as well:
$\displaystyle {du} = {\frac {1} {2 \sqrt {x + 1}} dx} $
$\displaystyle {du} = {\frac {1} {2 u} \ dx} $
The substitution for $ dx $ is:
$\displaystyle dx = 2u du $
Applying the substitution, including the infinitesimal, yields:
$\displaystyle {\int _{-1} ^{3} {\frac 1 2 e^{\sqrt{(x + 1)}}} dx} = {\int _{u(x = -1)} ^{u(x = 3)} {u e^u} du} $
$\displaystyle {\int _{-1} ^{3} {\frac 1 2 e^{\sqrt{(x + 1)}}} dx} = [(u - 1)e^u] _{u(x = -1)} ^{u(x = 3)} $
The integral can now be calculated (taking care to preserve the correct boundaries):
$\displaystyle u(x = -1) = \sqrt {-1 + 1} = 0 $
$\displaystyle u(x = 3) = \sqrt {3 + 1} = 2 $
$\displaystyle [(u - 1)e^u] _{u(x = -1)} ^{u(x = 3)} = (2 - 1)e^2 - (0 - 1)e^0  $
Thus, the solution for the formula:
$\displaystyle \int _{-1} ^{3} {\frac 1 2 e^{\sqrt{(x + 1)}}} dx = e^2 + 1 $
