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?
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 $
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 $