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evalute the integral
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\[\int\limits_{-\infty}^{-2}2dx/(x^2-1)\]
what does tht mean?
what u asked?
got ur question.haha..
ok wanna solve it?
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the indefinte integral is ln(x^2-1)
HINT ; whenever there is Infinty ... try to get rid of INfinty by Substituting 1/x as something
it is not ln.. coz on numerator there is no 2x
i cant solve it
in this case however partial fractionize the given ter m into (1/x-1) -(1/x+1)
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-ln(3)
ok ,got it now
what a easy question it was , i challenged myself ))
1.09861
\[2\int\limits_{-\infty}^{-2}dx\(x^2-1)\] \[2.1\2\left[ \log(-2-1)\(-2+1) \right] -2/2\left[ \log(-\infty-1)\(-\infty+1) \right]\] log3
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thanks to all contributors
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