Integral sqrt of (1+(x-1/(4x)^2) dx

[elcatracho]

Hey I've been working on this problem for awhile and I feel like i keep going in circles. I need to find the following:

. . .Integral of the square root of (1+(x-1/(4x)^2) dx

In words, that is the integral of the square root of one plus (x minus one over four x) squared. Thank you for your help.

 

[stapel]

Just to clarify: Is your integral as follows?

. . . . .\(\displaystyle \L \int{\,\sqrt{1\,+\,\left(\frac{x\,-\,1}{4x}\right)^2}\,dx\)

If so, then the answer is very messy.

Eliz.

 

[elcatracho]

no the x is seperate from the fraction. Let me try to rewrite it... it would be the square root of (1+ (x-((1)/(4x))^2) and I need to find the integral of that.

 

[stapel]

You have an unmatched parentheses, so I'm not sure. Do you mean the following?

. . . . .\(\displaystyle \L \int{\sqrt{1\,+\,\left(x\,-\,\frac{1}{4x}\right)^2}}\,dx\)

If so then the integral, while not as horrible as before, is still pretty messy.

Eliz.

 

[elcatracho]

yea, that is the integral i'm trying to find

 

[soroban]

Hello, elcatracho!

This is a truly awful problem!
I tried a variety of approaches ... and settled on Trig Substitution.

\(\displaystyle \L\int\sqrt{1\,+\,\left(x\,-\,\frac{1}{4x}\right)^2}\,dx\)

Let \(\displaystyle \,x\,-\,\frac{1}{4x}\:=\:\tan\theta\;\;\Rightarrow\;\;4x^2\,-\,4(\tan\theta)x\,-\,1\:=\:0\)

Quadratic Formula: \(\displaystyle \:x\:=\:\frac{4\tan\theta\,\pm\,\sqrt{16\tan^2\theta\,+\,16}}{8}\:=\:\frac{4\tan\theta\,\pm\,\sqrt{16(\tan^2\theta\,+\,1)}}{8}\:=\:\frac{4\tan\theta\,\pm\,\sqrt{16\sec^2\theta}}{8}\)

Then: \(\displaystyle \,x\:=\:\frac{4\tan\theta\,\pm\,4\sec\theta}{8}\:=\:\frac{1}{2}(\tan\theta\,\pm\,\sec\theta)\)

And: \(\displaystyle \:dx\:=\:\frac{1}{2}(\sec^2\theta\,\pm\,\sec\theta\tan\theta)\,d\theta\)

Substitute: \(\displaystyle \L\:\int\,\sec\theta\,\cdot\,\frac{1}{2}(\sec^2\theta\,\pm\,\sec\theta\tan\theta)\,d\theta\)

. . . . . \(\displaystyle \L=\;\frac{1}{2}\left[\int\sec^3\theta\,d\theta\:\pm\:\int\sec^2\theta\tan\theta\,d\theta\right]\)


The first integral can be solved by parts
. . or the formula: \(\displaystyle \:\int\sec x\,dx\:=\:\frac{1}{2}\left(\sec x\tan x\,+\,\ln|\sec x\,+\,\tan x| + C\)

The second can be solved with the substitution: \(\displaystyle \,u\,=\,\tan\theta\)


And we get: \(\displaystyle \L\:\frac{1}{4}\left[\sec\theta\tan\theta\,+\,\ln|\sec\theta\,+\,\tan\theta|\,\pm\,\tan^2\theta\right]\,+\,C\)

I'll let you back-substitute . . . I need a nap!

 

[elcatracho]

Thanks!

Thanks so much for your help!! I really appreciate your time and effort.

 

[skeeter]

hmm ... looks like a familiar arc-length problem that was carefully engineered to be easily integrated.

\(\displaystyle \L 1 + (x - \frac{1}{4x})^2 =\)

\(\displaystyle \L 1 + x^2 - \frac{1}{2} + \frac{1}{16x^2} =\)

\(\displaystyle \L x^2 + \frac{1}{2} + \frac{1}{16x^2} =\)

\(\displaystyle \L (x + \frac{1}{4x})^2\)

so the integral becomes ...

\(\displaystyle \L \int x + \frac{1}{4x} dx = \frac{x^2}{2} + \frac{\ln{|x|}}{4} + C\)