I'm having a mental lapse...
I need a hint with respect to which technique I am supposed to be using to integrate the following:
$\displaystyle \int{\sqrt{1+\frac{9}{4}x}}dx$
At one point I had the following:
$\displaystyle \int{\frac{\sqrt{9x+4}}{2}}dx$
What makes a good u substitution in this case? It seems like picking a value of $\displaystyle u = \sqrt{1+\frac{9}{4}x}$ does not seem to help. I hope I am missing something blatently obvious.
Yeah, I see now what I was doing wrong. Should have been:
$\displaystyle \int{1+(\frac{9}{4})x}dx$
let u = $\displaystyle 1+(\frac{9}{4})x$
$\displaystyle \int{u^\frac{1}{2}\frac{4}{9}du}$
$\displaystyle \frac{4}{9}\int{u^\frac{1}{2}du}$
$\displaystyle \frac{4}{9}\frac{2}{3}u^\frac{3}{2}+C$
$\displaystyle \frac{8}{27}u^\frac{3}{2}+C$
$\displaystyle \frac{8}{27}(1+(\frac{9}{4})x)^\frac{3}{2}+C$
$\displaystyle \frac{8}{27}(\frac{(9x+4)^\frac{3}{2}}{4^\frac{3}{ 2}}))+C$
$\displaystyle \frac{8}{27}(\frac{(9x+4)^\frac{3}{2}}{8}))+C$
$\displaystyle \frac{(9x+4)^\frac{3}{2}}{8}+C$
...
And this was actually a definite integral such that $\displaystyle 0\leq x \leq4$
So, I got a final answer of:
$\displaystyle \frac{80\sqrt{10}}{27}-\frac{8}{27}$
So...
$\displaystyle \frac{8(10\sqrt{10}-1)}{27}$