Question

How do you integrate $\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}$ using partial fractions?

Answer 1

Hint: We first try to describe the requirement and the process of finding the partial fractions. Then we factorise the denominator of $\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}$. We form the numerator of $\dfrac{5x-1}{{{x}^{2}}-x-2}$ with the factors. We simplify the equation $\dfrac{5x-1}{{{x}^{2}}-x-2}=\dfrac{A}{x-2}+\dfrac{B}{x+1}$ to find the partial fraction form.

Complete step by step solution:
For our given problem $\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}$, we first break ${{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1$ into factors.
We have ${{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1={{x}^{2}}\left( {{x}^{2}}+x-2 \right)+3{{x}^{2}}+1={{x}^{2}}\left( {{x}^{2}}+x-2 \right)+3\left( {{x}^{2}}+x-2 \right)-3x+7$.
 $\begin{align}
  & \dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2} \\
 & =\dfrac{{{x}^{2}}\left( {{x}^{2}}+x-2 \right)+3\left( {{x}^{2}}+x-2 \right)-3x+7}{{{x}^{2}}+x-2} \\
 & ={{x}^{2}}+3-\dfrac{3x-7}{{{x}^{2}}+x-2} \\
\end{align}$
Now we apply partial fractions for $\dfrac{3x-7}{{{x}^{2}}+x-2}$
Using middle-term factorisation we get
$\begin{align}
  & {{x}^{2}}+x-2 \\
 & ={{x}^{2}}+2x-x-2 \\
 & =x\left( x+2 \right)-\left( x+2 \right) \\
 & =\left( x+2 \right)\left( x-1 \right) \\
\end{align}$
Now we have to arrange the numerator $3x-7$ in the binary addition and subtraction of the same factors of ${{x}^{2}}+x-2$ which are $\left( x+2 \right),\left( x-1 \right)$.
We can do that directly or apply the form where $\dfrac{3x-7}{{{x}^{2}}+x-2}=\dfrac{A}{x+2}+\dfrac{B}{x-1}$.
We simplify the $\dfrac{A}{x+2}+\dfrac{B}{x-1}$. We get $\dfrac{A}{x+2}+\dfrac{B}{x-1}=\dfrac{A\left( x-1 \right)+B\left( x+2 \right)}{\left( x+2 \right)\left( x-1 \right)}$.
So, $\dfrac{3x-7}{{{x}^{2}}+x-2}=\dfrac{A}{x+2}+\dfrac{B}{x-1}=\dfrac{A\left( x-1 \right)+B\left( x+2 \right)}{\left( x+2 \right)\left( x-1 \right)}$ gives $3x-7=A\left( x-1 \right)+B\left( x+2 \right)$.
Simplifying we get $3x-7=A\left( x-1 \right)+B\left( x+2 \right)=x\left( A+B \right)+\left( 2B-A \right)$.
Equating the variables and constants we get $\left( A+B \right)=3;\left( 2B-A \right)=-7$.
Adding the equations, we get
$\begin{align}
  & \left( A+B \right)+\left( 2B-A \right)=3-7 \\
 & \Rightarrow 3B=-4 \\
 & \Rightarrow B=\dfrac{-4}{3} \\
\end{align}$
This gives $A=3-B=3-\dfrac{-4}{3}=\dfrac{13}{3}$.
The partial fraction form becomes \[\dfrac{3x-7}{{{x}^{2}}+x-2}=\dfrac{{}^{13}/{}_{3}}{x+2}+\dfrac{{}^{-4}/{}_{3}}{x-1}\].
The final form is $\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}={{x}^{2}}+3-\dfrac{{}^{13}/{}_{3}}{x+2}+\dfrac{{}^{4}/{}_{3}}{x-1}$
Now we integrate $\int{\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}dx}=\int{\left( {{x}^{2}}+3-\dfrac{{}^{13}/{}_{3}}{x+2}+\dfrac{{}^{4}/{}_{3}}{x-1} \right)dx}$.
\[\begin{align}
  & \int{\left( {{x}^{2}}+3-\dfrac{{}^{13}/{}_{3}}{x+2}+\dfrac{{}^{4}/{}_{3}}{x-1} \right)dx} \\
 & =\dfrac{{{x}^{3}}}{3}+3x-\dfrac{13}{3}\log \left| x+2 \right|+\dfrac{4}{3}\log \left| x-1 \right|+c \\
\end{align}\]
The integral form of $\dfrac{{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+1}{{{x}^{2}}+x-2}$ using partial fractions is \[\dfrac{{{x}^{3}}}{3}+3x-\dfrac{13}{3}\log \left| x+2 \right|+\dfrac{4}{3}\log \left| x-1 \right|+c\].

Note: The form gets trickier when there are the same factors within its power form. We can simplify the form by taking the form of $Ax+B$ instead of A in case of quadratics. These partial fraction forms are required for special forms of integration.