# Struggling with $\int \frac{dy}{y\left(1 - \frac y2\right)}$

I know I need to use a partial fraction and suspect I will end up with 2 terms that end up as a natural log integral but I just can't work it out. $$\int \frac{dy}{y\left(1 - \frac y2\right)}$$

I think that what I need to do is break it down into partial fractions or use substitutions after factoring :

For partial fractions I get to this point and cannot get any further

$z/(1/2 - (y+1)(y/2 + 1/2)$ <--- expanding out the original term

I am not sure how to turn this to decomposed partial fractions or how to deal with the -1/2

I expand out to get :

$1/(1/2 -(y+1)(y/2 + 1/2))$

$A/(y+1) + B/(y/2 + 1/2)$

I can't solve this.

$A(y/2 + 1/2) + B(y+1) =1$

For a substitution I get to this point :

$1/(1-u^2/2 du) where u = y + 1$

HINT : We have$$\frac{1}{y(1-(y/2))}=\frac{2}{y(2-y)}=\frac 1y+\frac{1}{2-y}=\frac 1y-\frac{1}{y-2}.$$

Hint :

The integrand can be decomposed as follows $$\frac{1}{y\left(1-\frac y2\right)}=\frac 1y+\frac{\frac12}{1-\frac y2}.$$

The idea is how to express $\displaystyle\frac{1}{y\left(1-\frac y2\right)}$ as $$\frac{A}{y}+\frac{B}{1-\frac y2}.$$ Now solve for $A$ and $B$ with cross multiplication and comparing the numerator RHS and LHS.

• so if i make y = 2 A drops off and B = 1/2 then make y 0 and a = 1 ? Sep 7, 2014 at 13:32
• @user3528592 If you cross multiply the last term, you will get $A+\left(B-\frac A2\right)y$ in the numerator. By comparing with the initial integrand, we will get $A=1$ and $B-\frac A2=0$. Sep 7, 2014 at 13:44

A similar approach:

$$\int \frac {dy}{y(1-\frac y2)}=2 \int \frac {dy}{y(2-y)}\\ =2\int \frac 12 \left[ \frac 1y +\frac 1{2-y} \right]dy\\ =\int \frac 1y +\frac 1{2-y}dy\\ =\ln y-\ln(2-y)\\ =\ln \frac y{2-y}$$