## anonymous 4 years ago 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

1. anonymous

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

2. anonymous

find n, rounded to the nearest whole

3. anonymous

$\int \frac{1}{x^2(1-x)} dx$ Hmm I would like to try partial fraction $\frac{1}{x^2(1+x)} = \frac{A}{1+x} + \frac{Bx +C}{x^2}$

4. anonymous

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

5. anonymous

$1 = Ax^2 + Bx^2 + Bx + Cx + C$ Comparing co-efficient C = 1; (A+B) = 0; (B+C) = 0 Now you can integrate it! :-D

6. anonymous

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.

7. anonymous

$\frac{1}{x^2(1+x)} = \frac{Ax^2 + Bx^2 + Bx + Cx + C}{x^2 (x+1)} \implies 1 = Ax^2 + Bx^2 + Bx + Cx + C$ This version is much better

8. anonymous

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

yeah exactly!

13. anonymous

my answer is 50 =) is that right?