Question

How do you integrate $\dfrac{3{{x}^{3}}-5{{x}^{2}}-11x+9}{{{x}^{2}}-2x-3}$ using partial fraction?

Answer 1

Hint: We know that we can convert a fraction into the sum of partial fractions if the degree of the numerator is less than the degree of the denominator. Also, we know that a non-repeated linear factor $s-a$ in the denominator, corresponds a partial fraction of the form $\dfrac{A}{s-a}.$

Complete step by step solution:
Let us consider the given fraction $\dfrac{3{{x}^{3}}-5{{x}^{2}}-11x+9}{{{x}^{2}}-2x-3}.$
We are asked to integrate this fraction using a partial fraction.
We know that a fraction can be converted to the sum of partial fractions if the degree of the numerator is less than that of the denominator.
Let us consider the numerator of the given fraction $3{{x}^{3}}-5{{x}^{2}}-11x+9.$
We will add and subtract ${{x}^{2}}$ and $x$ to this polynomial to get
\[\Rightarrow 3{{x}^{3}}-5{{x}^{2}}-11x+9=3{{x}^{3}}-5{{x}^{2}}+{{x}^{2}}-{{x}^{2}}+x-x-11x+9.\]
We will get
\[\Rightarrow 3{{x}^{3}}-5{{x}^{2}}-11x+9=3{{x}^{3}}-6{{x}^{2}}+{{x}^{2}}-x-10x+9=3{{x}^{3}}-6{{x}^{2}}-9x+{{x}^{2}}-2x+9.\]
When we take $3x$ from the first three summands we will get
\[\Rightarrow 3{{x}^{3}}-5{{x}^{2}}-11x+9=3x\left( {{x}^{2}}-2x-3 \right)+{{x}^{2}}-2x+9.\]
Let us add and subtract $3$ to get the equation as
\[\Rightarrow 3{{x}^{3}}-5{{x}^{2}}-11x+9=3x\left( {{x}^{2}}-2x-3 \right)+{{x}^{2}}-2x-3+9+3=3x\left( {{x}^{2}}-2x-3 \right)+\left( {{x}^{2}}-2x-3 \right)+12.\]
When we take the common out, we will get
\[\Rightarrow 3{{x}^{3}}-5{{x}^{2}}-11x+9=\left( 3x+1 \right)\left( {{x}^{2}}-2x-3 \right)+12.\]
Therefore, $\dfrac{3{{x}^{3}}-5{{x}^{2}}-11x+9}{{{x}^{2}}-2x-3}=3x+1+\dfrac{12}{{{x}^{2}}-2x-3}.$
We know that ${{x}^{2}}-2x-3=\left( x-3 \right)\left( x+1 \right).$
We will get $\dfrac{12}{{{x}^{2}}-2x-3}=\dfrac{12}{\left( x-3 \right)\left( x+1 \right)}=\dfrac{A}{x-3}+\dfrac{B}{x+1}.$
We will obtain $12=A\left( x+1 \right)+B\left( x-3 \right)=Ax+A+Bx-3B=\left( A+B \right)x+A-3B.$
Therefore, when we compare the coefficients, we will get $A+B=0$ and $A-3B=12.$
When we solve these equations, we will get $B=-A$ and so, $A-3B=A+3A=4A=12.$
That is, $A=\dfrac{12}{4}=3$ and $B=-3.$
Therefore, the sum of partial fraction of the given fraction is
$\Rightarrow \dfrac{3{{x}^{3}}-5{{x}^{2}}-11x+9}{{{x}^{2}}-2x-3}=3x+1+\dfrac{3}{x-3}-\dfrac{3}{x+1}.$

Note: To a repeated linear factor ${{\left( s-a \right)}^{r}}$ in the denominator, corresponds the sum of $r$ partial fractions of the form $\dfrac{{{A}_{1}}}{s-a}+\dfrac{{{A}_{2}}}{{{\left( s-a \right)}^{2}}}+\dfrac{{{A}_{3}}}{{{\left( s-a \right)}^{3}}}...+\dfrac{{{A}_{r}}}{{{\left( s-a \right)}^{r}}}.$ To a non-repeated quadratic factor ${{s}^{2}}+as+b$ in the denominator, corresponds a partial fraction of the form $\dfrac{As+B}{{{s}^{2}}+as+b}.$ To a repeated quadratic factor ${{\left( {{s}^{2}}+as+b \right)}^{r}}$ in the denominator, corresponds the sum of $r$ partial fractions of the form $\dfrac{{{A}_{1}}s+{{B}_{1}}}{{{s}^{2}}+as+b}+\dfrac{{{A}_{2}}s+{{B}_{2}}}{{{\left( {{s}^{2}}+as+b \right)}^{2}}}+...+\dfrac{{{A}_{r}}s+{{B}_{r}}}{{{\left( {{s}^{2}}+as+b \right)}^{r}}.}$