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integrate (1-x^2)/[x(1-2x)]dx
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\[\int\frac{1-x^2}{x(1-2x)}dx\] Have you tried partial fraction decomposition?
partial fraction cannot work, for that degree of numerator must be strictly less than degree of denom.
So long division first, then PFD.
yes.
I'd suggest rewriting as \[\int\frac{x^2-1}{2x^2-x}dx\] That should help with the long division
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k
x^2-1 = 1/2 (2x^2-2) = (1/2) (2x^2-x +x-2)
if you wantt o avoid long division....
got that ? ^ :P
hmmm..i have solveed it acc to 2 u @hartnn and i got 1/2[1+(1/2x-1)-(2/2x^2-x)]
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you got that as final answer ? and where is +c ?
nahi yaar till now i have only put value
yup i got the answer and have checked it from wolfram Thank You so much @hartnn
ohh...i was verifying manually :P welcome ^_^
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