# integrate!

• Dec 6th 2012, 12:15 AM
Tutu
integrate!
Hi I have problems integrating two questions, how do I integrate sqrt(x^2-4) / x? And (4lnx) / (x(1+(lnx)^2)? Thank you very much!!
• Dec 6th 2012, 04:33 AM
coolge
Re: integrate!
For the second problem, first use t= ln(x), then use z = 1+ t^2. Hope this helps!
• Dec 6th 2012, 08:36 AM
Soroban
Re: integrate!
Hello, Tutu!

Quote:

$\int\frac{\sqrt{x^2-4}}{x}\,dx$

It should be obvious that a Trig Substitution is called for.

Let $x \,=\,2\sec\theta \quad\Rightarrow\quad dx \,=\,2\sec\theta\tan\theta\,d\theta \quad\Rightarrow\quad \sqrt{x^2-4} \,=\,2\tan\theta$

Substitute: . $\int \frac{2\tan\theta}{2\sec\theta}(2\sec\theta\tan \theta\,d\theta) \;=\;2\int\tan^2\theta\,d\theta$

. . . . . . . . $=\;2\int(\sec^2-1)\,d\theta \;=\;2(\tan\theta - \theta) + C$

Now back-substitute.
• Dec 6th 2012, 08:38 AM
HallsofIvy
Re: integrate!
Quote:

Originally Posted by coolge
For the second problem, first use t= ln(x), then use z = 1+ t^2. Hope this helps!

Which is, of course, the same as the single substitution z= 1+ (ln(x))^2.
• Dec 6th 2012, 01:23 PM
topsquark
Re: integrate!
Quote:

Originally Posted by HallsofIvy
Which is, of course, the same as the single substitution z= 1+ (ln(x))^2.

My two cents. I prefer to use two separate substitutions rather than one more complicated one. I can keep better track that way. Or maybe I'm just a coward. (Sun)

-Dan