Under certain circumstances, the number of generations of "good" bacteria needed to increase the frequency of a particular gene from is n=4.572 integral 1/((x^2)(1-x)) dx

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Under certain circumstances, the number of generations of "good" bacteria needed to increase the frequency of a particular gene from is n=4.572 integral 1/((x^2)(1-x)) dx

Mathematics
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\[n=4.572\int\limits_{0.1}^{0.6}1/(x^2(1-x))dx\]
find n, rounded to the nearest whole
\[\int \frac{1}{x^2(1-x)} dx\] Hmm I would like to try partial fraction \[\frac{1}{x^2(1+x)} = \frac{A}{1+x} + \frac{Bx +C}{x^2}\]

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partial fraction been a long time... i think its the denominator your working on right?
\[1 = Ax^2 + Bx^2 + Bx + Cx + C \] Comparing co-efficient C = 1; (A+B) = 0; (B+C) = 0 Now you can integrate it! :-D
Umm the Numerator \[\frac{1}{x^2(1+x)} = \frac{Ax^2 + Bx^2 + Bx + Cx + C}{x^2 (x+1)} = 1 = Ax^2 + Bx^2 + Bx + Cx + C\] Denominators cancel out.
\[\frac{1}{x^2(1+x)} = \frac{Ax^2 + Bx^2 + Bx + Cx + C}{x^2 (x+1)} \implies 1 = Ax^2 + Bx^2 + Bx + Cx + C\] This version is much better
your 1/x^2(1-x) becomes 1/x^2(1+x) typo?
Oh Crap, yeah typo but the concept is same and I also messed up my calculations but the concept remains the same. Instead of \[\frac{1}{x^2(1+x)} = \frac{A}{1+x} + \frac{Bx +C}{x^2}\] you will have to do \[\frac{1}{x^2(1-x)} = \frac{A}{1-x} + \frac{Bx +C}{x^2}\]
so its gonna be 1=\[x^2(A-B)+x(B-C)+C\]
C= 1, (A-B) = 0, (B-C)=0
yeah exactly!
my answer is 50 =) is that right?

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