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Question: How do you integrate \[\int{\dfrac{{{x}^{4}}}{{{x}^{4}}-1}}\] using partial fractions?...

How do you integrate ∫x4x4−1\int{\dfrac{{{x}^{4}}}{{{x}^{4}}-1}} using partial fractions?

Explanation

Solution

For the given problem, we have to integrate ∫x4x4−1\int{\dfrac{{{x}^{4}}}{{{x}^{4}}-1}} using partial fractions. First of all we have to do a long division for partition. After that we have to assume an equation based on the denominator and then by using integration formulas we have to solve the remaining problem.

Complete step-by-step answer:
For the given question, we are given to integrate ∫x4x4−1\int{\dfrac{{{x}^{4}}}{{{x}^{4}}-1}} using partial fractions.
So, let us consider
I=∫x4x4−1.........(1)I=\int{\dfrac{{{x}^{4}}}{{{x}^{4}}-1}}.........\left( 1 \right)
Since the numerator and denominator have the same termx4{{x}^{4}}, we will add and subtract the numerator with 1.
I=∫x4−1+1x4−1I=\int{\dfrac{{{x}^{4}}-1+1}{{{x}^{4}}-1}}
By simplifying the equation, we get
=∫x4−1x4−1+1x4−1=\int{\dfrac{{{x}^{4}}-1}{{{x}^{4}}-1}+\dfrac{1}{{{x}^{4}}-1}}
=∫1+1x4−1=\int{1+\dfrac{1}{{{x}^{4}}-1}}
Let us consider
I=∫1+1x4−1..........(2)I=\int{1+\dfrac{1}{{{x}^{4}}-1}}..........(2)
Now we are comfortable to do a partial fraction.
Since the denominator is not a linear function, we need to check it if it is reducible.
x4−1=(x2+1)(x2−1){{x}^{4}}-1=\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}-1 \right)
Therefore, it is reducible
Since these functions are not linear functions. Let us check again whether these functions can be reduced or not.
x4−1=(x2+1)(x+1)(x−1){{x}^{4}}-1=\left( {{x}^{2}}+1 \right)\left( x+1 \right)\left( x-1 \right)
Therefore, again it is reduced.
Rewriting the equation (2) by replacing the above function.
I=∫1+1(x2+1)(x+1)(x−1)I=\int{1+\dfrac{1}{\left( {{x}^{2}}+1 \right)\left( x+1 \right)\left( x-1 \right)}}
Let us consider
I=∫1+ax+bx2+1+cx+1+dx−1.......(3)I=\int{1+\dfrac{ax+b}{{{x}^{2}}+1}+\dfrac{c}{x+1}+\dfrac{d}{x-1}}.......\left( 3 \right)
Now, we do partial fraction
1(x2+1)(x+1)(x−1)=ax+bx2+1+cx+1+dx−1\dfrac{1}{\left( {{x}^{2}}+1 \right)\left( x+1 \right)\left( x-1 \right)}=\dfrac{ax+b}{{{x}^{2}}+1}+\dfrac{c}{x+1}+\dfrac{d}{x-1}
1=Ax+b(x+1)(x−1)+c(x2+1)(x−1)+d(x2+1)(x+1)1=Ax+b\left( x+1 \right)\left( x-1 \right)+c\left( {{x}^{2}}+1 \right)\left( x-1 \right)+d\left( {{x}^{2}}+1 \right)\left( x+1 \right)
Let us consider
1=Ax+b(x+1)(x−1)+c(x2+1)(x−1)+d(x2+1)(x+1).........(4)1=Ax+b\left( x+1 \right)\left( x-1 \right)+c\left( {{x}^{2}}+1 \right)\left( x-1 \right)+d\left( {{x}^{2}}+1 \right)\left( x+1 \right).........\left( 4 \right)
For getting values of a, b, c, d let us substitute value in x.
Substitute x=1 in equation (4)
1=A(1)+b(1+1)(1−1)+c(1+1)(1−1)+d(1+1)(1+1)1=A\left( 1 \right)+b\left( 1+1 \right)\left( 1-1 \right)+c\left( 1+1 \right)\left( 1-1 \right)+d\left( 1+1 \right)\left( 1+1 \right)
1=4d1=4d
d=14d=\dfrac{1}{4}
Let us consider
d=14.........(5)d=\dfrac{1}{4}.........\left( 5 \right)
Substitute x=-1 in equation (4)
1=A(−1)+b((−1)+1)((−1)−1)+c((−1)2+1)((−1)−1)+d((−1)2+1)((−1)+1)1=A\left( -1 \right)+b\left( \left( -1 \right)+1 \right)\left( \left( -1 \right)-1 \right)+c\left( {{\left( -1 \right)}^{2}}+1 \right)\left( \left( -1 \right)-1 \right)+d\left( {{\left( -1 \right)}^{2}}+1 \right)\left( \left( -1 \right)+1 \right)

& 1=-4c \\\ & c=-\dfrac{-1}{4} \\\ \end{aligned}$$ Let us consider $$c=-\dfrac{-1}{4}...........\left( 6 \right)$$ Compare the coefficient of $${{x}^{3}}$$on both side $$0=a+c+d$$ From equation (5) and (6), we get $$0=a+\dfrac{1}{4}-\dfrac{-1}{4}$$ $$a=0$$ Let us consider $$a=0...........\left( 7 \right)$$ Comparing constant number on both side, we get $$\begin{aligned} & 1=-b-c+d \\\ & b=-1-c+d \\\ & \text{ =-1+}\dfrac{1}{4}+\dfrac{1}{4} \\\ & b=\dfrac{-1}{2} \\\ \end{aligned}$$ Let us consider $$b=\dfrac{-1}{2}..........\left( 8 \right)$$ Substituting a, b, c, d values in equation (3) $$I=\int{1+\dfrac{1}{\left( {{x}^{2}}+1 \right)\left( x+1 \right)\left( x-1 \right)}}=\int{1+\dfrac{\left( 0 \right)x+\left( \dfrac{-1}{2} \right)}{{{x}^{2}}+1}+\dfrac{\left( \dfrac{-1}{4} \right)}{x+1}+\dfrac{\left( \dfrac{1}{4} \right)}{x-1}}$$ $$\Rightarrow I=\int{1+\dfrac{-1}{2\left( {{x}^{2}}+1 \right)}+\dfrac{-1}{4\left( x+1 \right)}+\dfrac{1}{4\left( x-1 \right)}}$$ $$\Rightarrow I=\int{1-\dfrac{1}{2\left( {{x}^{2}}+1 \right)}-\dfrac{1}{4\left( x+1 \right)}+\dfrac{1}{4\left( x-1 \right)}}$$ $$\Rightarrow I=\int{1-\int{\dfrac{1}{2\left( {{x}^{2}}+1 \right)}}-\int{\dfrac{1}{4\left( x+1 \right)}}+\int{\dfrac{1}{4\left( x-1 \right)}}}$$ $$\Rightarrow I=x-\int{\dfrac{1}{2}.\dfrac{1}{\left( {{x}^{2}}+1 \right)}}-\int{\dfrac{1}{4}.\dfrac{1}{\left( x+1 \right)}}+\int{\dfrac{1}{4}.\dfrac{1}{\left( x-1 \right)}}$$ $$\Rightarrow I=x-\dfrac{1}{2}\int{\dfrac{1}{\left( {{x}^{2}}+1 \right)}}-\dfrac{1}{4}\int{.\dfrac{1}{\left( x+1 \right)}}+\dfrac{1}{4}\int{.\dfrac{1}{\left( x-1 \right)}}$$ Rewriting the equation using formulas: $$\int{\dfrac{1}{{{x}^{2}}+1}=\tan \left( x \right)}$$ and $$\int{\dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}=\ln \left( f\left( x \right) \right)}$$ $$\Rightarrow I=x-\dfrac{1}{2}\tan \left( x \right)-\dfrac{1}{4}\ln \left( x+1 \right)+\dfrac{1}{4}\ln \left( x-1 \right)$$ Rewriting the equation using formula $$\ln a-\ln b=\ln \dfrac{a}{b}$$ $$\Rightarrow I=x-\dfrac{1}{2}\tan \left( x \right)+\dfrac{1}{4}\ln \left( \dfrac{x-1}{x+1} \right)$$ Let us consider $$I=x-\dfrac{1}{2}\tan \left( x \right)+\dfrac{1}{4}\ln \left( \dfrac{x-1}{x+1} \right).......\left( 9 \right)$$ Therefore, integration of $$\int{\dfrac{{{x}^{4}}}{{{x}^{4}}-1}}$$ using partial fraction is $$I=x-\dfrac{1}{2}\tan \left( x \right)+\dfrac{1}{4}\ln \left( \dfrac{x-1}{x+1} \right)$$. **Note:** For solving this problem we should have a keen knowledge on partial fraction integration. Be careful while doing partition of a given fraction. Make sure that your fraction is completely linear. Students should have a keen knowledge on integration formulas for solving this kind of problems.