Question

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

Answer 1

Firstly use the partial fraction method to express the integrand in two separate integrands whose denominators are the individual factors of the given integrand. And then integrate them separately.
Partial fraction of the rational fraction form $\dfrac{{p(x) + q}}{{(x \pm a)(x \pm b)}}$ is given as $\dfrac{A}{{(x \pm a)}} + \dfrac{B}{{(x \pm b)}}$
Find the value of A and B by comparing it with original integrand.

Complete step by step solution:
In order to integrate $\dfrac{3}{{(x - 2)(x + 1)}}$ using partial fractions, we will first separate or express the integrand as sum of two integrands as follows
$\dfrac{3}{{(x - 2)(x + 1)}} = \dfrac{A}{{(x - 2)}} + \dfrac{B}{{(x + 1)}}$
Now, we will find the value of A and B by comparing both integrands as follows

$$3 = A(x + 1) + B(x - 2) \\\ \Rightarrow 3 = Ax + A + Bx - 2B \\\ \Rightarrow 3 = x(A + B) + (A - 2B) \\\$$

Equating similar terms (constants and coefficients of the variable) we will get,
$\Rightarrow A + B = 0\;and\;A - 2B = 3$
Subtracting the first equation from the second one, we will get

$$\;\;A - 2B = 3 \\\ \underline { - A - B = 0\;\;} \\\ \;\;\;\;\; - 3B = 3 \\\ \Rightarrow B = - 1 \\\$$

Putting value of B in first equation,
$A + ( - 1) = 0 \Rightarrow A = 1$
Now we have the respective values of A and B, so putting them in the above considered integrands and then integrating them, we will get
$\dfrac{3}{{(x - 2)(x + 1)}} = \dfrac{1}{{(x - 2)}} - \dfrac{1}{{(x + 1)}}$
Taking integration both sides, we will get
$\Rightarrow \int {\dfrac{3}{{(x - 2)(x + 1)}}dx} = \int {\left[ {\dfrac{1}{{(x - 2)}} - \dfrac{1}{{(x + 1)}}} \right]} dx$
Distributing the integration above subtraction, we will get
$\Rightarrow \int {\dfrac{3}{{(x - 2)(x + 1)}}dx} = \int {\dfrac{1}{{(x - 2)}} - \int {\dfrac{1}{{(x + 1)}}} } dx$
Now, we know that the integration of $\dfrac{1}{{(x \pm a)}}$ equals $\ln \left| {x \pm a} \right| + c$, so integrating the above functions using this formula, we will get
$\Rightarrow \int {\dfrac{3}{{(x - 2)(x + 1)}}dx} = \ln \left| {x - 2} \right| - \ln \left| {x + 1} \right| + c$
Using property of logarithm function, we can simplify it as follows
$\Rightarrow \int {\dfrac{3}{{(x - 2)(x + 1)}}dx} = \ln \left| {\dfrac{{x - 2}}{{x + 1}}} \right| + c$

Therefore $\ln \left| {\dfrac{{x - 2}}{{x + 1}}} \right| + c$ is the integration of $\dfrac{3}{{(x - 2)(x + 1)}}$

Note: When doing indefinite integration, always write $+ c$ part after the integration. This $+ c$ part indicates the constant part remains after integration and can be understood when you explore it graphically. In finite integration constant gets cancelled out, so we only write it in indefinite integration.