Question

How do you express \[\dfrac{3x}{{{x}^{2}}+x-2}\] in partial fractions?

Answer 1

Hint: In this question, the denominator is a quadratic equation that is ${{x}^{2}}+x-2$ and it cannot be expanded in real partial fraction. Therefore, to solve this we have to factor it into quadratic factors and then solve it in partial fraction decomposition.

Complete step by step solution:
We have our fraction as:
\[\Rightarrow \dfrac{3x}{{{x}^{2}}+x-2}\]
Since denominator is quadratic equation, we will factorize the denominator as:
$\Rightarrow {{x}^{2}}+x-2=\left( x+2 \right)\left( x-1 \right)$
Now, by applying in our fraction, we get:
\[\Rightarrow \dfrac{3x}{\left( x+2 \right)\left( x-1 \right)}\]
With this, we can do partial fraction decomposition in the form $\dfrac{A}{x+2}+\dfrac{B}{x-1}=\dfrac{3x}{\left( x+2 \right)\left( x-1 \right)}$
On simplifying left-hand side fraction that is making denominator equal, we get:
$\dfrac{A\left( x-1 \right)}{\left( x+2 \right)\left( x-1 \right)}+\dfrac{B\left( x+2 \right)}{\left( x+2 \right)\left( x-1 \right)}=\dfrac{3x}{\left( x+2 \right)\left( x-1 \right)}$
On simplifying left-hand side and right-hand side and cancelling terms, we get:
\[\Rightarrow Ax-A+Bx+2B=3x\]
Now we will write terms as similar terms.
Therefore, we get:
$\Rightarrow \left( A+B \right)x+\left( 2B-A \right)=3x$
With this, we can now write a system of equations.
Therefore, we get:
$A+B=3\to \left( 1 \right)$ and
$2B-A=0\to \left( 2 \right)$
On solving, equation $\left( 1 \right)$ , we get:
$\Rightarrow B=3-A$
Now on substituting $B=3-A$ in equation $\left( 2 \right)$ , we get:
$\Rightarrow 2\left( 3-A \right)-A=0$
On simplifying, we get:
$\Rightarrow 6-2A-A=0$
On further simplification, we get:
 $\Rightarrow -3A=-6$
On taking $3$ on right-hand side and cancelling minus sign, we get:
$\Rightarrow A=\dfrac{6}{3}$
On simplifying, we get:
$\Rightarrow A=2$
Now on substituting $A=2$ in equation $\left( 1 \right)$ , we get:
$A+B=3$
$2+B=3$
On simplifying, we get:
$B=3-2$
$B=1$
With this, now substituting our solution in actual practical fraction, we get our final answer that is:
$\Rightarrow \dfrac{2}{x+2}+\dfrac{1}{x-1}$
Therefore, the solution is:
$\Rightarrow \dfrac{2}{x+2}+\dfrac{1}{x-1}=\dfrac{3x}{\left( x+2 \right)\left( x-1 \right)}$

Note: It is to be remembered that while doing integration by parts the terms $u$ and $v$ should follow the sequence of the acronym $ILATE$, which stands for inverse, logarithm, algebraic, trigonometric and exponential respectively. partial fractions are used in questions of integration when we want to convert polynomial terms into linear terms.