Integration Question

**Ivanator27** · April 22nd 2012, 05:23 PM

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.

**Prove It** · April 22nd 2012, 06:41 PM

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.

$\displaystyle \begin{align*} \int{\frac{\sqrt{1 + 4ku\,t}}{2k\,t}\,dt} &= \frac{1}{2k}\int{\frac{\sqrt{1 + 4ku\,t}}{t}\,dt} \\ &= \frac{1}{2k}\int{\frac{1 + 4ku\,t}{t\sqrt{1 + 4ku\,t}}\,dt} \\ &= \frac{1}{4k^2u}\int{\frac{2ku\left(1 + 4ku\,t\right)}{t\sqrt{1 + 4ku\,t}}\,dt} \end{align*}$

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

**Ivanator27** · April 26th 2012, 12:15 AM

$\frac{1}{4k^2u}\int\frac{2ku(1+4kut)}{t\sqrt{1+4kut}}$

I agree with that, however once:

$v=\sqrt{1+4kut}$

$dv=\frac{2ku}{\sqrt{1+4kut}}dt$

$dt=\frac{\sqrt{1+4kut}}{2ku}dv$

Which leads me to:

$\frac{1}{4k^2u}\int\left ( \frac{2ku(v^2)}{t\sqrt{1+4kut}} \right )\left ( \frac{\sqrt{1+4kut}}{2ku} \right )dv$

$\frac{1}{4k^2u}\int\frac{v^2}{t}dv$

**biffboy** · April 26th 2012, 12:59 AM

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)

**biffboy** · April 26th 2012, 01:05 AM

End of my 1st line should have been 2kt=(z^2-1)/2u

**Ivanator27** · April 26th 2012, 01:24 AM

$z^{2}=1+4kut\Rightarrow 2zdz=4kudt$

$dt=\frac{2z}{4ku}dz$

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

$\frac{1}{4k^2u}\int\frac{z^2}{t}dz$

Which is exactly what happens if you let $z=\sqrt{1+4kut}$

Maybe I'm not understanding exactly what you were saying.

**biffboy** · April 26th 2012, 01:46 AM

It was just an alternative substitution. You now need to write t in terms of z.

**Ivanator27** · April 26th 2012, 02:12 AM

Ohhhh! Thanks so much! I have it know. =D

...Thanks again!

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