# Don't know where to start with this integral

• March 30th 2009, 11:23 AM
Stonehambey
Don't know where to start with this integral
I used to think I was pretty good at basic integration, then I met this

Show that

$\int^8_3\frac{(x+1)^{\frac{1}{2}}}{x+5} \, dx = \pi + 2 + \tan^{-1}\frac{3}{2}$

Any advice on how to even start this one would be much appreciated. I think I'm missing a trick somewhere... :)
• March 30th 2009, 11:35 AM
Jester
Quote:

Originally Posted by Stonehambey
I used to think I was pretty good at basic integration, then I met this

Show that

$\int^8_3\frac{(x+1)^{\frac{1}{2}}}{x+5} \, dx = \pi + 2 + \tan^{-1}\frac{3}{2}$

Any advice on how to even start this one would be much appreciated. I think I'm missing a trick somewhere... :)

Try letting $x + 1 = 4 \tan^2 \theta$.
• March 30th 2009, 11:58 AM
gammaman
It looks to me like it is coming from this formula

$\frac{1}{x^2+a^2}dx = \frac{1}{a}tan^{-1}\frac{x}{a}+c$
• March 30th 2009, 12:38 PM
TwistedOne151
Let $u=\sqrt{x+1}$, so that $x=u^2-1$, $dx=2u\,du$, and the integral becomes:
$\int_3^8\frac{(x+1)^{\frac{1}{2}}}{x+5}\,dx=\int_2 ^3\frac{u}{u^2-1+5}\cdot2u\,du=2\int_2^3\frac{u^2}{u^2+4}\,du$
$=2\int_2^3\frac{(u^2+4)-4}{u^2+4}\,du$
$=2\left[\int_2^3\,du-4\int_2^3\frac{1}{u^2+4}\,du\right]$
$=2-8\int_2^3\frac{1}{u^2+4}\,du$
Use the inverse tangent formula gammaman gave you for this last integral, and you should get the result, particularly if you recall that $\tan^{-1}(1)=\frac{\pi}{4}$.

--Kevin C.