1. ## Integration

Show that with limits 3 - 8, ∫(x+1)^1/2 / (x+5) dx = pi + 2 - 4 arctan 3/2.

How can I do this?

2. $\int_{3}^{8}\frac{\sqrt{x+1}}{x+5}~dx$

Let $t^2 = x+1$. Then $2t~dt = dx$.

$2\int_{2}^{3}\frac{t^2}{t^2+4}~dt$

To make it easier, I'll make one more substitution:

$2\int_{2}^{3}\frac{{\left(\frac{t}{2}\right)}^2}{{ \left(\frac{t}{2}\right)}^2+1}~dt$

$k = \frac{t}{2}$ and it becomes,

$4\int_{1}^{3/2}\frac{k^2}{k^2+1}~dk$

Now use integration by parts to get the answer.
$u=k$, $dv = \frac{k}{k^2+1}~dk$

There surely must be a shorter way. This is only a brute force attempt to get the integral. I'll think on it again as soon as I have some time (if someone doesn't help you in that time ;p)

3. Rewrite just a little.

$\frac{\sqrt{x+1}}{(x+1)+4}$

$\frac{Stuff}{Stuff^{2}+4}$

This should suggest a tantalizing substitution.