Improper Integral Trouble

[Sanguron]

The problem calls for me to integrate 1/(x^2 - 2x) from 8 to infinity. So, first I used partial fractions to break this into A/x + B/(x-2). Then I see that A(x-2) + Bx = 1, so A+B=0 and -2A=1. This tells me that A= -1/2 and B= 1/2.

then I integrate (-1/2x) + (1/2(x-2)). I took 1/2 out of both, leaving (-1/x) + (1/(x-2)), which can be integrated to get 1/2 ( -ln |-x| + ln |x-2| ) from 8 to infinity. The real answer is 1/2 ln3, but I get divergence. please help

 

[mmm4444bot]

… integrate 1/(x^2 - 2x) from 8 to infinity … The real answer is 1/2 ln(3) … This answer is incorrect. Please, double-check all of your typing.

 

[Sanguron]

sorry, it's from 3 to infinity

 

[BigGlenntheHeavy]

\(\displaystyle \int_{3}^{\infty} \frac{1}{x^{2}-2x}dx \ = \ \lim_{b\to\infty} \int_{3}^{b}\bigg[\frac{1}{2(x-2)}-\frac{1}{2x}\bigg]dx,\)

\(\displaystyle = \ \lim_{b\to\infty} \bigg[\frac{1}{2}ln|x-2|-\frac{1}{2}ln|x|\bigg]_{3}^{b} \ = \ \lim_{b\to\infty}\frac{1}{2}ln\bigg|\frac{x-2}{x}\bigg|_{3}^{b}\)

\(\displaystyle \mbox{the Marqui to the rescue:}\)

\(\displaystyle = \ \frac{1}{2}[ln(1)-ln(1/3)] \ = \ (1/2)[ln(3)] \ = \ \frac{ln(3)}{2}. \ QED.\)

 

[Sanguron]

thank you so much!