partial fractions

Integration Using Partial Fraction Decomposition

Partial fraction decomposition is the process of taking a rational function and decomposing it into simpler rational expressions which are easier to integrate.

We only use partial fractions if the rational function is proper. If

$f(x)=\frac{p(x)}{q(x)}$

then the degree of $p(x)$ must be less than the degree of $q(x)$ .

Types of Partial Fraction Decompositions

Factor	Term in Partial Fraction
$ax+b$	$\frac{A}{ax+b}$
$(ax+b)^n$	$\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} +\cdots+ \frac{A_n}{(ax+b)^n}$
Irreducible quadratic $ax^2+bx+c$	$\frac{Ax+B}{ax^2+bx+c}$

Example

$\int \frac{3x+5}{x^2-x-2}\,dx$

Find

Factorise the denominator:

$x^2-x-2=(x-2)(x+1)$

Write

$\frac{3x+5}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$

Multiplying through by

$(x-2)(x+1)$ :

$3x+5=A(x+1)+B(x-2)$

Let $x=-1$ :

$2=-3B$

$B=-\frac23$

Let $x=2$ :

$11=3A$

$A=\frac{11}{3}$

Hence

$\frac{3x+5}{x^2-x-2} = \frac{11}{3(x-2)} - \frac{2}{3(x+1)}$

Integrating:

$\int\frac{3x+5}{x^2-x-2}\,dx = \int \left( \frac{11}{3(x-2)} - \frac{2}{3(x+1)} \right) dx$

$= \frac{11}{3}\ln|x-2| -\frac23\ln|x+1| +C$

Example

Find

$\int\frac{3x+2}{(x-1)^2}\,dx$

Write

$\frac{3x+2}{(x-1)^2} = \frac{A_1}{x-1} + \frac{A_2}{(x-1)^2}$

Multiplying through by $(x-1)^2$ :

$3x+2=A_1(x-1)+A_2$

Let $x=1$ :

$5=A_2$

Let $x=0$ :

$2=-A_1+5$

$A_1=3$

Therefore

$\int\frac{3x+2}{(x-1)^2}\,dx = \int \left( \frac3{x-1} + \frac5{(x-1)^2} \right) dx$

$= 3\ln|x-1| -\frac5{x-1} +C$

Example

Find

$\int \frac{2x-5} {(x-1)(x^2+1)} \,dx$

Write

$\frac{2x-5} {(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$

Multiplying through:

$2x-5 = A(x^2+1) + (Bx+C)(x-1)$

Let $x=1$ :

$-3=2A$

$A=-\frac32$

Let $x=0$ :

$-5=-\frac32-C$

$C=\frac72$

Let $x=-1$ :

$-7=-3+\left(-B+\frac72\right)(-2)$

$-4=2B-7$

$B=\frac32$

Hence

$\int \frac{2x-5} {(x-1)(x^2+1)} \,dx = \frac12 \int \left( -\frac3{x-1} + \frac{3x}{x^2+1} + \frac7{x^2+1} \right) dx$

$= \frac12 \left( -3\ln|x-1| +\frac32\ln(x^2+1) +7\arctan(x) \right) +C$

$= -\frac32\ln|x-1| +\frac34\ln(x^2+1) +\frac72\arctan(x) +C$

The logistic differential equation

$\begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}$

where $r$ is the growth parameter and $k$ is the carrying capacity.

And the maximum rate of increase happens when $P=\frac{k}{2}$

$\begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}$

$\begin{equation*}\frac{dP}{P(k-P)}=r dt{\end{equation}$

$\begin{equation*}\int \frac{dP}{P(k-P)}=\int r dt{\end{equation}$

I am going to separate the denominator on the left hand side

\frac{1}{P(k-P)}=\frac{A}{P}+\frac{B}{k-P}

\frac{1}{P(k-P)}=\frac{A(k-P)+BP}{P(k-P)}

So our equation is,

$\begin{equation*}\int \frac{\frac{1}{k}}{P}+\frac{\frac{1}{k}}{k-P} dP=\int r dt\end{equation}$

$\begin{equation*}\frac{1}{k}\int \frac{1}{P}+\frac{1}{k-P} dP=\int r dt\end{equation}$

$\begin{equation*}\int \frac{1}{P}+\frac{1}{k-P} dP=\int kr dt\end{equation}$

$\begin{equation*}ln\lvert{P}\rvert-ln\lvert{k-P}\rvert=krt+c\end{equation}$

$\begin{equation*}ln\lvert{\frac{P}{k-P}\rvert=krt+c\end{equation}$

$\begin{equation*}\frac{P}{k-P}=e^{krt+c}\end{equation}$

$\begin{equation*}\frac{P}{k-P}=e^{krt}e^{c} \end{equation}$

When $t=0, P=P_0$ ,

$\begin{equation*}\frac{P_0}{k-P_0}=e^{c} \end{equation}$

The equation is now

$\begin{equation*}\frac{P}{k-P}=\frac{P_0}{k-P_0}e^{krt}\end{equation}$

$\begin{equation*}P=\frac{P_0}{k-P_0}e^{krt}(k-P)\end{equation}$

$\begin{equation*}P=k\frac{P_0}{k-P_0}e^{krt}-P\frac{P_0}{k-P_0}e^{krt}\end{equation}$

$\begin{equation*}P+P\frac{P_0}{k-P_0}e^{krt}=k\frac{P_0}{k-P_0}e^{krt}\end{equation}$

$\begin{equation*}P(1+\frac{P_0}{k-P_0}e^{krt})=k\frac{P_0}{k-P_0}e^{krt}\end{equation}$

$\begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{1+\frac{P_0}{k-P_0}e^{krt}}\end{equation}$

$\begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{\frac{k-P_0+P_0e^{krt}}{k-P_0}}\end{equation}$

$\begin{equation*}P=\frac{kP_0e^{rkt}}{k-P_0+P_0e^{rkt}}\end{equation}$

Divide by $e^{rkt}$

$\begin{equation*}P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}$

$\begin{equation*}}\frac{dP}{dt}=rP(k-P)\Longleftrightarrow P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}$

Proving the Maximum Rate of Increase Happens When $P=\frac{k}{2}$

$\begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}$

$\begin{equation*}\frac{d^2P}{dt^2}=r\frac{dP}{dt}(k-P)+rP(-\frac{dP}{dt})\end{equation}$

$\begin{equation*}\frac{d^2P}{dt^2}=\frac{dP}{dt}(rk-rP-rP)\end{equation}$

$\begin{equation*}\frac{d^2P}{dt^2}=0\end{equation}$

$\begin{equation*}\frac{dP}{dt}(rk-rP-rP)=0\end{equation}$

$\begin{equation*}r\frac{dP}{dt}(k-2P)=0\end{equation}$

$\begin{equation*}\frac{dP}{dt}(k-2P)=0\end{equation}$

$\begin{equation*}rP(k-P)(k-2P)=0\end{equation}$

Hence $P=k$ or $P=\frac{k}{2}$

(1) $\begin{equation*}\frac{d^3P}{dt^3}=\frac{dP^2}{dt^2}(rk-2rP)+\frac{dP}{dt}(-2\frac{dP}{dt})\end{equation*}$

Substitute $P=k$ into equation $1$

$\begin{equation*}\frac{d^3P}{dt^3}=rk(k-k)(rk-2rk)(rk-2rk)-2(rk(k-k))^2=0\end{equation}$

Hence, not a maximum.

Substitute $P=\frac{k}{2}$ into equation $1$

$\begin{equation*}\frac{d^3P}{dt^3}=rk(k-\frac{k}{2})(rk-2r\frac{k}{2})(rk-2r\frac{k}{2})-2(rk(k-\frac{k}{2}))^2=0\end{equation}$

$\begin{equation*}\frac{d^3P}{dt^3}=-2(rk^2-\frac{rk^2}{2})^2\end{equation}$

$\begin{equation*}\frac{d^3P}{dt^3}=-2\frac{r^2k^4}{4}\end{equation}$

$-2\frac{r^2k^4}{4}\le 0$ For all values of $P, r$ and $k$ .

Hence maximum when $P=\frac{k}{2}$

We will look at a worked example in the next post.

Tag Archives: partial fractions

Integration Using Partial Fraction Decomposition

Example

Example

Example

Deriving the Logistic Growth Equation

Proving the Maximum Rate of Increase Happens When $P=\frac{k}{2}$

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