Hi all,
Could someone please show me how to integrate:
[ ( 1 + 4kut ) ^1/2 ] / 2kt dt
I've tried everything, maybe I'm just being dim but I can't find a substitution that works and I can't do it by parts.
$\displaystyle \displaystyle \begin{align*} \int{\frac{\sqrt{1 + 4ku\,t}}{2k\,t}\,dt} &= \frac{1}{2k}\int{\frac{\sqrt{1 + 4ku\,t}}{t}\,dt} \\ &= \frac{1}{2k}\int{\frac{1 + 4ku\,t}{t\sqrt{1 + 4ku\,t}}\,dt} \\ &= \frac{1}{4k^2u}\int{\frac{2ku\left(1 + 4ku\,t\right)}{t\sqrt{1 + 4ku\,t}}\,dt} \end{align*}$
Let $\displaystyle \displaystyle \begin{align*} v = \sqrt{1 + 4ku\,t} \implies dv = \frac{2ku}{\sqrt{1 + 4ku\,t}}\,dt \end{align*}$ and the integral becomes
$\displaystyle \displaystyle \begin{align*} \frac{1}{4k^2u}\int{\frac{v^2}{\frac{v^2 - 1}{4ku}}\,dv} &= \frac{1}{4k^2u}\int{\frac{4ku\,v^2}{v^2 - 1}\,dv} \\ &= \frac{1}{k}\int{\frac{v^2}{v^2 - 1}\,dv} \\ &= \frac{1}{k}\int{\frac{v^2 - 1 + 1}{v^2 - 1}\,dv} \\ &= \frac{1}{k}\int{1 + \frac{1}{v^2 - 1}\,dv}\end{align*}$
You can now integrate this using partial fractions