Determine if the improper integral is divergent or not

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Fatima Hasan
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Homework Statement


Determine if the improper integral is divergent or convergent .
Improper integral from 0 to 2 of x/(x^2 - 1)^2 dx, with the integrand written as x over (x squared minus 1) squared.

Homework Equations


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The Attempt at a Solution


Handwritten-style typeset calculation of the improper integral of x/(x^2 - 1)^2 from 0 to 2, split at x = 1 with limits, with a red line showing the antiderivative evaluated at the limits giving -∞ + ∞ = 0, and a final result of -1/6 + ∞.

When i solved the first term using online calculator , the answer was "The integral is divergent" . However , I got 0 .
Where is my mistake ?
 
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  • Improper integral from 0 to 2 of x/(x^2 - 1)^2 dx, with the integrand written as x over (x squared minus 1) squared.
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  • Handwritten-style typeset calculation of the improper integral of x/(x^2 - 1)^2 from 0 to 2, split at x = 1 with limits, with a red line showing the antiderivative evaluated at the limits giving -∞ + ∞ = 0, and a final result of -1/6 + ∞.
    screenshot_32.png
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You cannot generally do things like ##\infty - \infty = 0##. Furthermore, what you have is not ##\infty - \infty## as ##1/[2(0^2 - 1)] = -1/2##.
 
Orodruin said:
You cannot generally do things like ##\infty - \infty = 0##. Furthermore, what you have is not ##\infty - \infty## as ##1/[2(0^2 - 1)] = -1/2##.
It should be ##-1/[2(1^2-1)] +- 1/[2(0^2-1)]= -1/0 + 1/2 = -\infty+1/2##
So , it's divergent . Right ?