Question

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

Answer 1

Hint: First we expand the denominator according to $\left( {a + b} \right)\left( {a - b} \right)$. By simplifying and solving further, we cancel out the like denominators. Then, we equate the coefficients of $x$ and coefficients of the constants on both sides. After we have found the values of $A$ and $B$, we integrate our given number.

Complete step by step answer:
In the given question, we need to integrate the expression $\dfrac{{2x + 3}}{{{x^2} - 9}}$ using partial fraction. We find that the given term is in the form of proper fraction as the power on the denominator is more than that on the numerator.
Now we know that: ${a^2} - {b^2}$ can also be expressed as $\left( {a + b} \right)\left( {a - b} \right)$
Hence the denominator can also be written as: $\left( {{x^2} - {3^2}} \right) = \left( {x + 3} \right)\left( {x - 3} \right)$
Now, let $\dfrac{{2x + 3}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{A}{{\left( {x + 3} \right)}} + \dfrac{B}{{\left( {x - 3} \right)}}$
On simplifying the right hand side further, we find:
$ \Rightarrow \dfrac{{2x + 3}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{A\left( {x - 3} \right) + B\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}$
We cancel out the denominators from both the sides, we get:
$ \Rightarrow 2x + 3 = A\left( {x - 3} \right) + B\left( {x + 3} \right)$
Simplifying the right hand side further by opening the brackets and multiplying, we get:
$ \Rightarrow 2x + 3 = Ax - 3A + Bx + 3B$
We group the constant terms and the coefficient terms together:
$ \Rightarrow 2x + 3 = \left( {A + B} \right)x - 3\left( {A - B} \right)$
Equating the coefficients of $x$ and coefficients of the constants on both sides, we get:
$ \Rightarrow 2x = \left( {A + B} \right)x$ ……………. EQUATION (1)
and $3 = - 3\left( {A - B} \right)$ …………….EQUATION (2)
On solving equation 1, we get:
$ \Rightarrow 2 = A + B$
Therefore, $A = 2 - B$
Substituting the above value in EQUATION (2), we get:
$ \Rightarrow - 3\left( {A - B} \right) = 3$
$ \Rightarrow - 3\left( {2 - B - B} \right) = 3$
On simplifying it further, we get:
$ \Rightarrow - 2 + 2B = 1$
On adding $2$ to both sides, we get:
$ \Rightarrow 2B = 1 + 2$
$ \Rightarrow 2B = 3$
On simplifying even further, we get:
$B = \dfrac{3}{2}$
Since $A = 2 - B$
Therefore, $A = 2 - \dfrac{3}{2} = \dfrac{1}{2}$
As we know that $\dfrac{{2x + 3}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{A}{{\left( {x + 3} \right)}} + \dfrac{B}{{\left( {x - 3} \right)}}$
Placing the values of $A$ and $B$ , in the above equation, we get:
$ \Rightarrow \dfrac{{2x + 3}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{\dfrac{1}{2}}}{{\left( {x + 3} \right)}} + \dfrac{{\dfrac{3}{2}}}{{\left( {x - 3} \right)}}$
Let’s integrate them:
$ \Rightarrow \int {\dfrac{{2x + 3}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}dx = \int {\dfrac{{\dfrac{1}{2}}}{{\left( {x + 3} \right)}} + \dfrac{{\dfrac{3}{2}}}{{\left( {x - 3} \right)}}dx} } $
$ \Rightarrow \int {\dfrac{{2x + 3}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}dx = \int {\dfrac{{\dfrac{1}{2}}}{{\left( {x + 3} \right)}}dx + \int {\dfrac{{\dfrac{3}{2}}}{{\left( {x - 3} \right)}}dx} } } $
Taking the constant terms outside the integration sign:
$ \Rightarrow \int {\dfrac{{2x + 3}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}dx = \dfrac{1}{2}\int {\dfrac{{dx}}{{\left( {x + 3} \right)}} + \dfrac{3}{2}\int {\dfrac{{dx}}{{\left( {x - 3} \right)}}} } } $
As we know that the integration of $\int {\dfrac{{dx}}{x} = \log \left| x \right| + C} $
Therefore, $ \Rightarrow \int {\dfrac{{2x + 3}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}dx = \dfrac{1}{2}\log \left| {x + 3} \right|} + \dfrac{3}{2}\log \left| {x - 3} \right| + C$
Thus, we have our required answer.

Note: Integration is a type of calculus along with differentiation. Integration simply means to add up smaller parts of any area, volume, etc to give us the whole value. There are different types of integration methods to solve more complex multiplication and division questions like:
Integration by substitution
Integration by parts, etc.