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Question: How do you evaluate the integral \(\int {\dfrac{{dx}}{{{x^4} - 16}}} \)?...

How do you evaluate the integral ∫dxx4−16\int {\dfrac{{dx}}{{{x^4} - 16}}} ?

Explanation

Solution

We will first use the identity of a2−b2{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 ∫dxx4−16\int {\dfrac{{dx}}{{{x^4} - 16}}} .
It has the denominator of x4−16{x^4} - 16.
Using the fact that: a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b), we will get x4−16=(x2−4)(x2+4){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:-
⇒x4−16=(x2−4)(x2+4)=(x−2)(x+2)(x2+4)\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:-
⇒1x4−16=1(x−2)(x+2)(x2+4)\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:-
⇒1x4−16=Ax+2+Bx−2+Cx+Dx2+4\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:-
⇒1=A(x−2)(x2+4)+B(x+2)(x2+4)+(Cx+D)(x−2)(x+2)\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:-
⇒1=A(x3−2x2+4x−8)+B(x3+2x2+4x+8)+(Cx+D)(x2−4)\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:-
⇒1=(A+B+C)x3+(−2A+2B+D)x2+(4A+4B−4C)x+(−8A+8B−4D)\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:-
⇒A+B+C=0\Rightarrow A + B + C = 0 …………(1)
⇒−2A+2B+D=0\Rightarrow - 2A + 2B + D = 0 …………(2)
⇒4A+4B−4C=0\Rightarrow 4A + 4B - 4C = 0 …………(3)
⇒−8A+8B−4D=1\Rightarrow - 8A + 8B - 4D = 1 …………(4)
Solving the equation numbers from (1) to (4), we will get:-
⇒A=−132,B=132,C=0\Rightarrow A = - \dfrac{1}{{32}},B = \dfrac{1}{{32}},C = 0 and D=−18D = - \dfrac{1}{8}
Putting these values in the equation (A), we will get:-
⇒1x4−16=−132(x+2)+132(x−2)−18(x2+4)\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:-
⇒∫dxx4−16=−132∫dxx+2+132∫dxx−2−18∫dx(x2+4)\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 ∫dxx+a=ln⁡∣x+a∣+C\int {\dfrac{{dx}}{{x + a}} = \ln |x + a| + C} . So, we will get:-
⇒∫dxx4−16=−132ln⁡∣x+2∣+132ln⁡∣x−2∣−18∫dx(x2+4)\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 ∫dx(x2+a2)=tan⁡−1xa+C\int {\dfrac{{dx}}{{\left( {{x^2} + {a^2}} \right)}} = {{\tan }^{ - 1}}\dfrac{x}{a} + C} . So, we get:-
⇒∫dxx4−16=−132ln⁡∣x+2∣+132ln⁡∣x−2∣−18tan⁡−1x2+C\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 −132ln⁡∣x+2∣+132ln⁡∣x−2∣−18tan⁡−1x2+C - \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 ⇒∫dxx4−16=−132∫dxx+2+132∫dxx−2−18∫dx(x2+4)\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 ⇒1x4−16=−132(x+2)+132(x−2)−18(x2+4) \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: ∫(f+g+h)(x)dx=∫f(x)dx+∫g(x)dx+∫h(x)dx\int {(f + g + h)(x)dx = \int {f(x)dx} } + \int {g(x)dx} + \int {h(x)dx} , where we took: f(x)=1x+2,g(x)=1x−2f(x) = \dfrac{1}{{x + 2}},g(x) = \dfrac{1}{{x - 2}} and h(x)=1x2+4h(x) = \dfrac{1}{{{x^2} + 4}} and thus, we have the result stated above.
The students must commit to the memory the following formulas:-
a2−b2=(a−b)(a+b){a^2} - {b^2} = (a - b)(a + b)
∫dx(x2+a2)=tan⁡−1xa+C\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.