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How to integrate (2u+1)/(2-2u^2) du?
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\[\frac{ 2u+1 }{ 2-2u^2 }du\]
factor out the 2 first
numerator = u+1 + u
denominator = 2(1-u)(1+u)
So (1/2) times the integral of (2u+1)/(1-u^2) du?
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1/2(1-u) + u/2(1-u^2)
mow integrate all this separately. for second integral put 1-u^2 =t and this can be solved.
Will partial fractions work?
i don't think they are required nut yes they can be applied in order to get the answer.
**but
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Okay, thanks!
ok i am happy that i cud help u
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