Question

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

Answer 1

In this question we have to integrate the given function using partial fractions, this can be done by first factor the denominator into linear factors, write the partial fraction for each factor, then multiply the whole equation by the denominator, then equate the new expression with the given function, and find the values of the coefficients, and then by using the identity i.e.,$\int {\dfrac{1}{{x + n}}dx = \ln \left| {x + n} \right|}$, we will integrate the expression.

Complete step-by-step solution:
Given function is $\dfrac{{2x + 3}}{{{x^2} - 9}}$,
Since the numerator is in a lower degree than the denominator, we can proceed to the next step,
Now we can write the function as,
\Rightarrow $$$$\int {\dfrac{{2x + 3}}{{{x^2} - 9}}dx},
Now factor the denominator we get,
\Rightarrow $$$$\int {\dfrac{{2x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}dx},
Now rewriting the function in partial fraction we get,
$\Rightarrow \dfrac{A}{{x + 3}} + \dfrac{B}{{x - 3}} = \dfrac{{2x + 3}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}$,
Now taking L.C.M we get,
$\Rightarrow \dfrac{{A\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} + \dfrac{{B\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{2x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}$,
Now simplifying we get,
$\Rightarrow \dfrac{{Ax - 3A}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} + \dfrac{{Bx + 3B}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{{2x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}$,
Now adding the terms we get,
$\Rightarrow \dfrac{{Ax - 3A + Bx + 3B}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{2x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}$,
Now separating the $x$ terms and coefficients terms we get,
$\Rightarrow \dfrac{{x\left( {A + B} \right) + 3\left( {B - A} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{2x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}$,
Comparing the two functions we get,
$A + B = 2 - - - - (1)$ and
$B - A = 1 - - - - (2)$,
Now eliminating the like terms we get,
$\Rightarrow 2B = 3$,
Now dividing 2 to both sides we get,
$\Rightarrow \dfrac{{2B}}{2} = \dfrac{3}{2}$,
Now simplifying we get,
$\Rightarrow B = \dfrac{3}{2}$,
Now substituting the value of B in (2) we get,
$\Rightarrow \dfrac{3}{2} - A = 1$,
Now subtracting$\dfrac{3}{2}$ to both sides we get,
$\Rightarrow \dfrac{3}{2} - A - \dfrac{3}{2} = 1 - \dfrac{3}{2}$,
Now simplifying we get,
$\Rightarrow - A = \dfrac{{ - 1}}{2}$,
Now taking out the negative sign we get,
$\Rightarrow A = \dfrac{1}{2}$,
Now substituting the values in the original form we get,
$\Rightarrow \int {\dfrac{1}{2} \cdot \dfrac{1}{{x + 3}} + \dfrac{3}{2} \cdot \dfrac{1}{{x - 3}}dx}$,
Now separating the integration we get,
$\Rightarrow \int {\dfrac{1}{2} \cdot \dfrac{1}{{x + 3}}dx + \int {\dfrac{3}{2} \cdot \dfrac{1}{{x - 3}}} dx}$,
Now taking out the constants from integration we get,
$\Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{x + 3}}dx + \dfrac{3}{2}\int {\dfrac{1}{{x - 3}}} dx}$,
Now applying integration using the identity i.e.,$\int {\dfrac{1}{{x + n}}dx = \ln \left| {x + n} \right|}$, we get,
$\Rightarrow \dfrac{1}{2}\ln \left| {x + 3} \right| + \dfrac{3}{2}\ln \left| {x - 3} \right|$,
Now further simplification we get,
$\Rightarrow \dfrac{{\ln \left| {x + 3} \right| + 3\ln \left| {x - 3} \right|}}{2} + C$,
So, the integration value for the given function is $\dfrac{{\ln \left| {x + 3} \right| + 3\ln \left| {x - 3} \right|}}{2} + C$.

$\therefore$The integrated value when the given function $\dfrac{{2x + 3}}{{{x^2} - 9}}$ is integrated using partial fractions will be equal to $\dfrac{{\ln \left| {x + 3} \right| + 3\ln \left| {x - 3} \right|}}{2} + C$.

Note: The Partial fractions can be used in Mathematics to turn functions that cannot be integrated into simple fractions easily. We can basically use partial fractions if the degree of the numerator is strictly less than the degree of the denominator.