Question

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

Answer 1

\[n=4.572\int\limits_{0.1}^{0.6}1/(x^2(1-x))dx\]

Answer 2

find n, rounded to the nearest whole

Answer 3

\[\int \frac{1}{x^2(1-x)} dx\]
Hmm I would like to try partial fraction

\[\frac{1}{x^2(1+x)} = \frac{A}{1+x} + \frac{Bx +C}{x^2}\]

Answer 4

partial fraction been a long time... i think its the denominator your working on right?

Answer 5

\[1 = Ax^2 + Bx^2 + Bx + Cx + C \]
Comparing co-efficient C = 1; (A+B) = 0; (B+C) = 0

Now you can integrate it! :-D

Answer 6

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.

Answer 7

\[\frac{1}{x^2(1+x)} = \frac{Ax^2 + Bx^2 + Bx + Cx + C}{x^2 (x+1)} \implies 1 = Ax^2 + Bx^2 + Bx + Cx + C\]

This version is much better

Answer 8

your 1/x^2(1-x) becomes 1/x^2(1+x) typo?

Answer 9

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 \[\frac{1}{x^2(1-x)} = \frac{A}{1-x} + \frac{Bx +C}{x^2}\]

Answer 10

so its gonna be 1=\[x^2(A-B)+x(B-C)+C\]

Answer 11

C= 1, (A-B) = 0, (B-C)=0

Answer 12

yeah exactly!

Answer 13

my answer is 50 =) is that right?