-1
$\begingroup$

I've been googling some tutorials on integrating logarithms for my calc 2 class and I've found a lot of good stuff. Unfortunately nothing has answered how to handle a problem that I have. I've tried u-substitution and have still been getting it wrong or just lost. My problem is the $\int(\frac{9}{x^3} + \frac{2}{9x})dx$ (sorry, I'm a little new to the math section and don't know how to format that nicely). So far I've tried finding a common denominator which changes the equation to $\frac{81+2x^2}{9x^3}$. I tried doing $u = 9x^3$ and $du = 27x^2dx$ or $\frac{1}{27}du = x^2dx$. Now the problem is kind of ugly and I'm not sure what to do with that next. We haven't covered integration by parts yet in this class and I've found that comes up, so I'm not sure if it's possible to evaluate without that technique. I really appreciate any help!

Thank you in advance, Frank

$\endgroup$
4
$\begingroup$

There is no need to use substitution in this case. This is pretty straight forward and can be calculated directly using the formula.

$\int (\frac{9}{x^3}+\frac{2}{9x})dx$

$=9\int x^{-3}dx+\frac{2}{9}\int\frac{dx}{x}$

$=9(\frac{x^{-3+1}}{-3+1})+\frac{2}{9}\ln x+c$

$=-\frac{9}{2}\frac{1}{x^2}+\frac{2}{9}\ln x+c$

In case you don't know which formulae have been used, see here :

$\int x^ndx=\frac{x^{n+1}}{n+1}+c, -1\ne n\in \mathbb{Z}$
$\int\frac{dx}{x}=\ln x+c$. This is a special case of the previous formula for $n=-1$.

For a list of all basic integral formulae see here.

$\endgroup$
  • 1
    $\begingroup$ Wow thank you so much! I was way over thinking that. I was so used to trying to cram ln|x| into the whole thing that I forgot to rewrite it a simpler way before trying u substitution, or not realizing I didn't need to do that -_- Thank you so much for the help! $\endgroup$ – Frank A. Sep 8 '14 at 6:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.