Question

How do you evaluate the integral $\int {\dfrac{{dx}}{{{x^4} - 16}}} $?

Answer 1

Hint: We will first use the identity of ${a^2} - {b^2}$ and then use the method of partial fraction to find it in easier terms, which we can integrate easily and get the answer.

Complete step-by-step answer:
We are given that we are required to find the value of $\int {\dfrac{{dx}}{{{x^4} - 16}}} $.
It has the denominator of \[{x^4} - 16\].
Using the fact that: ${a^2} - {b^2} = (a - b)(a + b)$, we will get \[{x^4} - 16 = \left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)\]
Now, using the same identity further, we will get the following expression:-
\[ \Rightarrow {x^4} - 16 = \left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right) = \left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 4} \right)\]
Now, we can write:-
\[ \Rightarrow \dfrac{1}{{{x^4} - 16}} = \dfrac{1}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 4} \right)}}\]
Using the partial fraction, we can write it as:-
\[ \Rightarrow \dfrac{1}{{{x^4} - 16}} = \dfrac{A}{{x + 2}} + \dfrac{B}{{x - 2}} + \dfrac{{Cx + D}}{{{x^2} + 4}}\] ………………….(A)
Simplifying it, we will get the following expression:-
\[ \Rightarrow 1 = A(x - 2)({x^2} + 4) + B(x + 2)({x^2} + 4) + \left( {Cx + D} \right)(x - 2)(x + 2)\]
Simplifying the right hand side, by opening the brackets, we will get:-
\[ \Rightarrow 1 = A\left( {{x^3} - 2{x^2} + 4x - 8} \right) + B\left( {{x^3} + 2{x^2} + 4x + 8} \right) + \left( {Cx + D} \right)\left( {{x^2} - 4} \right)\]
Combining the coefficients of x raised to the power 3, 2, 1 and 0 together to get the following expression:-
\[ \Rightarrow 1 = \left( {A + B + C} \right){x^3} + \left( { - 2A + 2B + D} \right){x^2} + \left( {4A + 4B - 4C} \right)x + \left( { - 8A + 8B - 4D} \right)\]
Comparing both the sides in the above equation, we will then obtain the following equation:-
\[ \Rightarrow A + B + C = 0\] …………(1)
\[ \Rightarrow - 2A + 2B + D = 0\] …………(2)
\[ \Rightarrow 4A + 4B - 4C = 0\] …………(3)
\[ \Rightarrow - 8A + 8B - 4D = 1\] …………(4)
Solving the equation numbers from (1) to (4), we will get:-
\[ \Rightarrow A = - \dfrac{1}{{32}},B = \dfrac{1}{{32}},C = 0\] and \[D = - \dfrac{1}{8}\]
Putting these values in the equation (A), we will get:-
\[ \Rightarrow \dfrac{1}{{{x^4} - 16}} = - \dfrac{1}{{32\left( {x + 2} \right)}} + \dfrac{1}{{32\left( {x - 2} \right)}} - \dfrac{1}{{8\left( {{x^2} + 4} \right)}}\]
Now, taking integration on both sides, we have:-
\[ \Rightarrow \int {\dfrac{{dx}}{{{x^4} - 16}}} = - \dfrac{1}{{32}}\int {\dfrac{{dx}}{{x + 2}}} + \dfrac{1}{{32}}\int {\dfrac{{dx}}{{x - 2}}} - \dfrac{1}{8}\int {\dfrac{{dx}}{{\left( {{x^2} + 4} \right)}}} \]
We know that $\int {\dfrac{{dx}}{{x + a}} = \ln |x + a| + C} $. So, we will get:-
\[ \Rightarrow \int {\dfrac{{dx}}{{{x^4} - 16}}} = - \dfrac{1}{{32}}\ln |x + 2| + \dfrac{1}{{32}}\ln |x - 2| - \dfrac{1}{8}\int {\dfrac{{dx}}{{\left( {{x^2} + 4} \right)}}} \]
Now, we also know that \[\int {\dfrac{{dx}}{{\left( {{x^2} + {a^2}} \right)}} = {{\tan }^{ - 1}}\dfrac{x}{a} + C} \]. So, we get:-
\[ \Rightarrow \int {\dfrac{{dx}}{{{x^4} - 16}}} = - \dfrac{1}{{32}}\ln |x + 2| + \dfrac{1}{{32}}\ln |x - 2| - \dfrac{1}{8}{\tan ^{ - 1}}\dfrac{x}{2} + C\]

Hence, the answer is \[ - \dfrac{1}{{32}}\ln |x + 2| + \dfrac{1}{{32}}\ln |x - 2| - \dfrac{1}{8}{\tan ^{ - 1}}\dfrac{x}{2} + C\].

Note:
The students must note that in the solution, when we got \[ \Rightarrow \int {\dfrac{{dx}}{{{x^4} - 16}}} = - \dfrac{1}{{32}}\int {\dfrac{{dx}}{{x + 2}}} + \dfrac{1}{{32}}\int {\dfrac{{dx}}{{x - 2}}} - \dfrac{1}{8}\int {\dfrac{{dx}}{{\left( {{x^2} + 4} \right)}}} \] from \[ \Rightarrow \dfrac{1}{{{x^4} - 16}} = - \dfrac{1}{{32\left( {x + 2} \right)}} + \dfrac{1}{{32\left( {x - 2} \right)}} - \dfrac{1}{{8\left( {{x^2} + 4} \right)}}\], we did not directly get it.
But, we used the fact that: $\int {(f + g + h)(x)dx = \int {f(x)dx} } + \int {g(x)dx} + \int {h(x)dx} $, where we took: $f(x) = \dfrac{1}{{x + 2}},g(x) = \dfrac{1}{{x - 2}}$ and $h(x) = \dfrac{1}{{{x^2} + 4}}$ and thus, we have the result stated above.
The students must commit to the memory the following formulas:-
 ${a^2} - {b^2} = (a - b)(a + b)$
\[\int {\dfrac{{dx}}{{\left( {{x^2} + {a^2}} \right)}} = {{\tan }^{ - 1}}\dfrac{x}{a} + C} \]
The students must know that, in partial fraction, we basically rewrite the function in such a manner that it becomes easier and simpler for us to find the integration of those functions using simpler formulas.