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 :) $$ I =\frac{\sqrt 2}{51}(\pi -2 \arctan(\sqrt 2)) + \frac4{51}\ln(3) - \frac7{51} \ln(2) $$
$\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 $x=3$