Question

How do you integrate $\dfrac{{2x}}{{({x^2} - 25)}}$ using partial fractions?

Answer 1

Firstly factorize the denominator of the integrand and use a 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 $\dfrac{{2x}}{{({x^2} - 25)}}$ using partial fractions, we will first factorize the denominator of the given expression
$= {x^2} - 25$
Using algebraic identity ${a^2} - {b^2} = (a + b)(a - b)$ to factorize it,
$= {x^2} - {5^2} \\\ = (x + 5)(x - 5) \\\$
Now, expressing the integrand as sum of two integrands as follows
$\dfrac{{2x}}{{(x + 5)(x - 5)}} = \dfrac{A}{{(x + 2)}} + \dfrac{B}{{(x - 5)}}$
Now, we will find the value of A and B by comparing both integrands as follows

$$2x = A(x - 5) + B(x + 5) \\\ \Rightarrow 2x = Ax - 5A + Bx + 5B \\\ \Rightarrow 2x = x(A + B) + 5(B - A) \\\$$

Equating similar terms (constants and coefficients of the variable) we will get,
$\Rightarrow A + B = 2\;and\;B - A = 0$
Adding both equations, we will get

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

Putting value of B in first equation,
$A + 1 = 2 \Rightarrow A = 1$
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
$\dfrac{{2x}}{{(x + 5)(x - 5)}} = \dfrac{1}{{(x + 5)}} + \dfrac{1}{{(x - 5)}}$
Taking integration both sides, we will get
$\Rightarrow \int {\dfrac{{2x}}{{(x + 5)(x - 5)}}dx} = \int {\left[ {\dfrac{1}{{(x + 5)}} + \dfrac{1}{{(x - 5)}}} \right]} dx$
Distributing the integration above subtraction, we will get
$\Rightarrow \int {\dfrac{{2x}}{{(x + 5)(x - 5)}}dx} = \int {\dfrac{1}{{(x + 5)}} + \int {\dfrac{1}{{(x - 5)}}} } dx$
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
$\Rightarrow \int {\dfrac{{2x}}{{(x + 5)(x - 5)}}dx} = \ln \left| {x + 5} \right| + \ln \left| {x - 5} \right| + c$
Using property of logarithm function, we can simplify it as follows
$\Rightarrow \int {\dfrac{{2x}}{{(x + 5)(x - 5)}}dx} = \ln \left| {(x + 5)(x - 5)} \right| + c \\\ \Rightarrow \int {\dfrac{{2x}}{{(x + 5)(x - 5)}}dx} = \ln \left| {{x^2} - {5^2}} \right| + c \\\ \Rightarrow \int {\dfrac{{2x}}{{(x + 5)(x - 5)}}dx} = \ln \left| {{x^2} - 25} \right| + c \\\$

Therefore $\ln \left| {{x^2} - 25} \right| + c$ is the integration of $\dfrac{{2x}}{{(x + 5)(x - 5)}}$

Note:
When expressing the integrand in partial fraction form, take care of the signs and also cross check it once by adding or subtracting the partial fractions, if you are getting the original integrand or not, because sometimes students calculate the incorrect partial fractions of the integrand and get incorrect answer after integrating them.