# Integration Question

• Apr 22nd 2012, 06:23 PM
Ivanator27
Integration Question
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.
(Doh)
• Apr 22nd 2012, 07:41 PM
Prove It
Re: Integration Question
Quote:

Originally Posted by Ivanator27
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.
(Doh)

\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 \begin{align*} v = \sqrt{1 + 4ku\,t} \implies dv = \frac{2ku}{\sqrt{1 + 4ku\,t}}\,dt \end{align*} and the integral becomes

\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 :)
• Apr 26th 2012, 01:15 AM
Ivanator27
Re: Integration Question
• Apr 26th 2012, 01:59 AM
biffboy
Re: Integration Question
Let z^2=1+4kut SO 2zdz/dt=4ku So zdz/2ku z^2-1=4kut So 2kt=(z^2-1)/u

Integral becomes 1/2k integral of z^2/(z^2-1) Now write z^2/(z^2-1)= 1+1/(z^2-1) and partial fraction 1/(z^2-1)
• Apr 26th 2012, 02:05 AM
biffboy
Re: Integration Question
End of my 1st line should have been 2kt=(z^2-1)/2u
• Apr 26th 2012, 02:24 AM
Ivanator27
Re: Integration Question
http://latex.codecogs.com/gif.latex?...w%202zdz=4kudt

http://latex.codecogs.com/gif.latex?dt=\frac{2z}{4ku}dz

I don't see how this helps me, it leads to:

http://latex.codecogs.com/gif.latex?...frac{z^2}{t}dz

Which is exactly what happens if you let http://latex.codecogs.com/gif.latex?z=\sqrt{1+4kut}

Maybe I'm not understanding exactly what you were saying.
• Apr 26th 2012, 02:46 AM
biffboy
Re: Integration Question
It was just an alternative substitution. You now need to write t in terms of z.
• Apr 26th 2012, 03:12 AM
Ivanator27
Re: Integration Question
Ohhhh! Thanks so much! I have it know. =D

...Thanks again!