How to integrate int dx/(x^5sqrt.(4x^2-1))

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I can't figure out what method to use here. Once I know the method, I can do the rest. I tried trig substitution but it got much more complicated than I thought it should have. Here's the problem.Can someone help? Thanks,

integral dx/(x^5sqrt.(4x^2-1))

If this is an Arcsec problem, I don't know how to change the x^5 in the denominator to equal u.
 
They can be tricky.

Try \(\displaystyle \L\\x=\frac{1}{2}sec({\theta}) \;\ and \;\ dx=\frac{1}{2}sec({\theta})tan({\theta})d{\theta}\)

\(\displaystyle \L\\\int\frac{1}{x^{5}\sqrt{4(x^{2}-\frac{1}{4})}}dx\)

After you sub, you get it whittled down to(verify):

\(\displaystyle \L\\\int\frac{\frac{1}{2}sec({\theta})tan({\theta})}{\frac{1}{32}sec^{5}({\theta})\sqrt{4(\frac{1}{4}(sec^{2}({\theta})-1))}}d{\theta}\)

\(\displaystyle \L\\16\int\frac{1}{sec^{4}({\theta})}d{\theta}=16\int{cos^{4}({\theta})d{\theta}\)

Now, you can finish?.
 
I can definitely solve from that point. However, Why factor a 4 out under the sqrt. Doesn't x=sec2theta in this integral? Where do you get that 1/2sectheta?
 
I just done that so you could see how we got to \(\displaystyle sec^{2}({\theta})-1=tan^{2}({\theta})\)

I edited the post. Take another gander. I hope it helps.
 
You could also use u-substitution:

\(\displaystyle \L\\\int\frac{x}{x^{6}\sqrt{4x^{2}-1}}dx\)

Let \(\displaystyle \L\\u=4x^{2}-1;\;\ \frac{du}{8}=xdx; \;\ x^{2}=\frac{u+1}{4}\)
 
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