Show that with limits 3 - 8, ∫(x+1)^1/2 / (x+5) dx = pi + 2 - 4 arctan 3/2.
How can I do this?
$\displaystyle \int_{3}^{8}\frac{\sqrt{x+1}}{x+5}~dx$
Let $\displaystyle t^2 = x+1$. Then $\displaystyle 2t~dt = dx$.
$\displaystyle 2\int_{2}^{3}\frac{t^2}{t^2+4}~dt$
To make it easier, I'll make one more substitution:
$\displaystyle 2\int_{2}^{3}\frac{{\left(\frac{t}{2}\right)}^2}{{ \left(\frac{t}{2}\right)}^2+1}~dt$
$\displaystyle k = \frac{t}{2}$ and it becomes,
$\displaystyle 4\int_{1}^{3/2}\frac{k^2}{k^2+1}~dk$
Now use integration by parts to get the answer.
$\displaystyle u=k$, $\displaystyle dv = \frac{k}{k^2+1}~dk$
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)