# This Improper integral triggers me

I'm an IT engineering student that for the past 20 minutes tried to solve this improper integral without succeeding .. Could you help me, please? The integral is kind of an easy one: $$I =\int_3^{+\infty} \frac{2x-6}{(2x^2+9)(x^2 -2x)} \,dx$$

After some calculus I've found that the indefinite integral is $$\frac 1{51}(17\ln|x| -3\ln|x-2| - 7\ln(2x^2 +9) +2 \sqrt 2 \arctan(\frac{\sqrt 2}3x) + C$$ Now I should continue with the $$\lim_{x\to +\infty} f(x)$$ but I find some difficulties.. This is the solution: $$I =\frac{\sqrt 2}{51}(\pi -2 \arctan(\sqrt 2)) + \frac4{51}\ln(3) - \frac7{51} \ln(2)$$ I don't know where I get wrong because if I do the limit, I get $$\lim_{x\to +\infty} \frac 1{51}(17\ln|x| -3\ln|x-2| - 7\ln(2x^2 +9) +2 \sqrt 2 \arctan(\frac{\sqrt 2}3x) = \frac{\sqrt 2\pi}{51}$$ "because arctan is faster than the natural logarithm"... But maybe it's here my mistake.. I KNOW for sure that I've made a mistake in this limit... I feel it. I haven't considered something. I've done so many limits that when I get wrong one, I know it ahah.. Thanks in advance guys! Enjoy :)

EDIT AFTER SOLUTION: Basically thanks to Doug M and the other members, I understood that I had done a STUPID mistake in the limit and so : $$\lim_{x\to +\infty} (17\ln|x| -3\ln|x-2| - 7\ln(2x^2 +9) =$$ $$\lim_{x\to +\infty} ln|\frac{x^{17}}{(x-2)^3(2x^2 +9)^7}|\simeq ln|\frac{x^{17}}{(2)^7(x)^{17} + O(x^{17})}|\simeq ln|\frac1{(2)^7}|\simeq -7ln(2)$$

So the correct whole limit is: $$\lim_{x\to +\infty} \frac 1{51}(17\ln|x| -3\ln|x-2| - 7\ln(2x^2 +9) +2 \sqrt 2 \arctan(\frac{\sqrt 2}3x) = \frac1{51}(\sqrt 2\pi - 7ln(2))$$ In order to continue this triggering integral (because if you start something in math, you have to see it through), I need to evaluate the expression at x = 3 and I get: $$\frac 1{51}(-4\ln(3) +2 \sqrt 2 \arctan(\sqrt 2))$$ And so FINALLY the integral becomes: $$I =\frac1{51}(\sqrt 2\pi - 7ln(2)) - [\frac 1{51}(-4\ln(3) +2 \sqrt 2 \arctan(\sqrt 2))]$$ And I got the solution :) $$I =\frac{\sqrt 2}{51}(\pi -2 \arctan(\sqrt 2)) + \frac4{51}\ln(3) - \frac7{51} \ln(2)$$

• This is a nice first question. Welcome to the site! I imagine "$arctg$" is in fact "$\arctan$"? – The Count Jan 19 '17 at 2:42
• It looks like you have correctly evaluated what is happening at the upper bound, but you also need to evaluate the lower bound. That should introduce the missing terms. – Doug M Jan 19 '17 at 2:44
• Love the title, made me laugh – Teh Rod Jan 19 '17 at 3:00
• @TheCount ... yes, "tan" (US) = "tg" (much of Europe). – GEdgar Jan 19 '17 at 3:05

$\lim_\limits{x\to +\infty}$$\frac 1{51}\bigg(17\ln|x| -3\ln|x-2| - 7\ln(2x^2 +9) +2 \sqrt 2 \arctan(\frac{\sqrt 2}3x)\bigg)\\ \frac 1{51}\bigg(\ln |\dfrac {x^{17}}{(x-2)^3(2x^2+9)^7}| +2 \sqrt 2 \arctan(\frac{\sqrt 2}3x)\bigg)$
In that logarithm we have $x^{17}$ and $2^7x^{17}$ as the highest powered terms of $x$ in the numerator and the denominator. As $x$ goes to infinity.
$\frac 1{51}(-7\ln 2 + \sqrt 2 \pi)$
And you still need to evaluate the expression at $x=3$
• I put in a little more detail. You expand out the polynomials. As $x$ gets to be very large, the highest powered terms are going to dominate. So, you don't really need to do the full expansion, you just need to know the power and the coefficients of the highest powered terms. – Doug M Jan 19 '17 at 2:58