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Question: How do you evaluate the integral \[\int {\dfrac{{{x^4}}}{{{x^2} - 1}}dx?} \]...

How do you evaluate the integral ∫x4x2−1dx?\int {\dfrac{{{x^4}}}{{{x^2} - 1}}dx?}

Explanation

Solution

We use the concepts of integration to solve this problem. In order to solve this, first of all we will simplify the given function to get it into an integrable form. And then we will further simply by using the partial fraction method. After that we will use the formulas of integration to get the required result.
Formula used:
∫xndx=xn+1n+1+c\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c}
∫1xdx=ln⁡∣x∣+c\int {\dfrac{1}{x}dx = \ln |x| + c}
where cc is the constant of integration

Complete step by step answer:
We have to evaluate the integral ∫x4x2−1dx\int {\dfrac{{{x^4}}}{{{x^2} - 1}}dx}
Let us consider the given integral as,
I=∫x4x2−1dx −−−(i)I = \int {\dfrac{{{x^4}}}{{{x^2} - 1}}dx} {\text{ }} - - - \left( i \right)
First of all, we will simplify the integrand term to get it into an integrable form
So, for the integrand x4x2−1\dfrac{{{x^4}}}{{{x^2} - 1}}
On adding and subtracting 11 in the numerator we get
x4+1−1x2−1\dfrac{{{x^4} + 1 - 1}}{{{x^2} - 1}}
⇒(x2)2−1+1x2−1 −−−(ii)\Rightarrow \dfrac{{{{\left( {{x^2}} \right)}^2} - 1 + 1}}{{{x^2} - 1}}{\text{ }} - - - \left( {ii} \right)
We know that
(a2−b2)=(a+b)(a−b)\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)
Therefore, from the equation (ii)\left( {ii} \right) we get
⇒(x2−1)(x2+1)+1x2−1\Rightarrow \dfrac{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) + 1}}{{{x^2} - 1}}
Now on separating the denominator, we get
⇒(x2−1)(x2+1)x2−1+1x2−1\Rightarrow \dfrac{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}{{{x^2} - 1}} + \dfrac{1}{{{x^2} - 1}}
⇒(x2+1)+1x2−1\Rightarrow \left( {{x^2} + 1} \right) + \dfrac{1}{{{x^2} - 1}}
⇒(x2+1)+1(x−1)(x+1) −−−(iii)\Rightarrow \left( {{x^2} + 1} \right) + \dfrac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}{\text{ }} - - - \left( {iii} \right)
Now we perform a partial fraction decomposition on the 2nd term in the equation (iii)\left( {iii} \right)
The second term is: 1(x−1)(x+1)\dfrac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}
Using partial fraction, we can write
1(x−1)(x+1)=A(x−1)+B(x+1) −−−(Z)\dfrac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{A}{{\left( {x - 1} \right)}} + \dfrac{B}{{\left( {x + 1} \right)}}{\text{ }} - - - \left( Z \right)
On multiplying by (x−1)(x+1)\left( {x - 1} \right)\left( {x + 1} \right) on both sides, we get
1=A(x−1)(x+1)(x−1)+B(x−1)(x+1)(x+1)1 = \dfrac{{A\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)}} + \dfrac{{B\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x + 1} \right)}}
On cancelling, we get
1=A(x+1)+B(x−1)1 = A\left( {x + 1} \right) + B\left( {x - 1} \right)
⇒1=Ax+A+Bx−B\Rightarrow 1 = Ax + A + Bx - B
On combining the like terms, we get
⇒1=x(A+B)+(A−B)\Rightarrow 1 = x\left( {A + B} \right) + \left( {A - B} \right)
On comparing both sides, we get
A+B=0−−−(a)A + B = 0 - - - \left( a \right)
A−B=1−−−(b)A - B = 1 - - - \left( b \right)
On solving equation (a)\left( a \right) and (b)\left( b \right) we get
A=12, B=−12A = \dfrac{1}{2},{\text{ }}B = \dfrac{{ - 1}}{2}
Therefore, on putting the values in the equation (Z)\left( Z \right) we get
1(x−1)(x+1)=12(x−1)−12(x+1)\dfrac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{1}{{2\left( {x - 1} \right)}} - \dfrac{1}{{2\left( {x + 1} \right)}}
Now on putting in the equation (iii)\left( {iii} \right) we get
⇒(x2+1)+12(x−1)−12(x+1)\Rightarrow \left( {{x^2} + 1} \right) + \dfrac{1}{{2\left( {x - 1} \right)}} - \dfrac{1}{{2\left( {x + 1} \right)}}
Therefore, we get our integrand as
x4x2−1=(x2+1)+12(x−1)−12(x+1)\dfrac{{{x^4}}}{{{x^2} - 1}} = \left( {{x^2} + 1} \right) + \dfrac{1}{{2\left( {x - 1} \right)}} - \dfrac{1}{{2\left( {x + 1} \right)}}
So, from equation (i)\left( i \right) we get
I=∫(x2+1)+12(x−1)−12(x+1) dxI = \int {\left( {{x^2} + 1} \right) + \dfrac{1}{{2\left( {x - 1} \right)}} - \dfrac{1}{{2\left( {x + 1} \right)}}} {\text{ }}dx
Now we will integrate the above equation,
We know that
∫xndx=xn+1n+1+c\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c}
∫1xdx=ln⁡∣x∣+c\int {\dfrac{1}{x}dx = \ln |x| + c}
Therefore, we get
I=(x33+x) + 12ln⁡∣x−1∣ − 12ln⁡∣x+1∣ +cI = \left( {\dfrac{{{x^3}}}{3} + x} \right){\text{ }} + {\text{ }}\dfrac{1}{2}\ln |x - 1|{\text{ }} - {\text{ }}\dfrac{1}{2}\ln |x + 1|{\text{ }} + c
We know that
log⁡a−log⁡b=log⁡ab\log a - \log b = \log \dfrac{a}{b}
Therefore, we get
I=(x33+x) + 12ln⁡∣x−1∣∣x+1∣ +cI = \left( {\dfrac{{{x^3}}}{3} + x} \right){\text{ }} + {\text{ }}\dfrac{1}{2}\ln \dfrac{{|x - 1|}}{{|x + 1|}}{\text{ }} + c
Hence, we get the required result.

Note:
The given integral is an example of an indefinite integral. So, after integrating and finding an indefinite integral, make sure that you add an arbitrary constant cc . One more thing to note is that after calculating partial fractions, once again simplify them and check whether you are getting the same result from which you found the partial fractions.