Question

How do you integrate $\dfrac{1}{y\left( 1-y \right)}$ ?

Answer 1

Hint: We are given a function in y-variable which is a product of two functions in the denominator in the y-variable. Integration of this quantity would be very difficult. Thus, we shall break this single term down into two parts as sum or difference of two terms which are easily integrable. Then we will integrate the modified equation using basic rules of integration.

Complete step-by-step solution:
We are given the equation $\dfrac{1}{y\left( 1-y \right)}$.
Taking the negative sign common from the function in the denominator, this can also be written as $\dfrac{-1}{y\left( y-1 \right)}$.
On further modification, we get $-\dfrac{\left( y-\left( y-1 \right) \right)}{y\left( y-1 \right)}$.
We shall now break this term into two parts as follows.
$\Rightarrow -\dfrac{\left( y-\left( y-1 \right) \right)}{y\left( y-1 \right)}=-\left( \dfrac{y}{y\left( y-1 \right)}-\dfrac{y-1}{y\left( y-1 \right)} \right)$
Cancelling the same terms from the numerator and denominator in both terms, we get
$\Rightarrow -\dfrac{\left( y-\left( y-1 \right) \right)}{y\left( y-1 \right)}=-\left( \dfrac{1}{y-1}-\dfrac{1}{y} \right)$
$\Rightarrow \dfrac{1}{y\left( 1-y \right)}=-\left( \dfrac{1}{y-1}-\dfrac{1}{y} \right)$
Now, we will simply integrate this equation.
\[\begin{align}
  & \Rightarrow \int{\dfrac{1}{y\left( 1-y \right)}}.dy=\int{-\left( \dfrac{1}{y-1}-\dfrac{1}{y} \right).dy} \\
 & \Rightarrow \int{\dfrac{1}{y\left( 1-y \right)}}.dy=-\int{\left( \dfrac{1}{y-1}-\dfrac{1}{y} \right).dy} \\
 & \Rightarrow \int{\dfrac{1}{y\left( 1-y \right)}}.dy=-\left[ \int{\dfrac{1}{y-1}.dy-\int{\dfrac{1}{y}.dy}} \right] \\
 & \Rightarrow \int{\dfrac{1}{y\left( 1-y \right)}}.dy=\int{\dfrac{1}{y}.dy}-\int{\dfrac{1}{y-1}.dy} \\
\end{align}\]
Using the property of integration, $\int{\dfrac{1}{x}.dx=\ln x+C}$, we get
\[\Rightarrow \int{\dfrac{1}{y\left( 1-y \right)}}.dy=\ln y-\ln \left( y-1 \right)+C\]
Where $C$ is the constant of integration.
Using logarithmic property, $\ln a-\ln b=\ln \dfrac{a}{b}$, we can further simplify it as:
\[\Rightarrow \int{\dfrac{1}{y\left( 1-y \right)}}.dy=\ln \dfrac{y}{\left( y-1 \right)}+C\]
Therefore, the integral of $\dfrac{1}{y\left( 1-y \right)}$ is \[\ln \dfrac{y}{\left( y-1 \right)}+C\].

Note: The constant of integration represents indefinite integration of a function. This is when the limits of integration are not specified. Another method of solving this problem is by applying integration by parts. This is done by using the ILATE rule which is used when the product of two separately integrable functions is given. We would have first brought the terms in the denominator to the numerator by raising them to a power of -1 and then applied the ILATE rule.