Question

How do you integrate \[\int {\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)}}dx} \] using partial fractions ?

Answer 1

Hint: In this question, we will try to represent the complex expression in the simplified manner in the combination of fractions which has the same value and then we will integrate the partial fractions separately to get the required solution. Partial fraction is the decomposition of an equation which is of the form \[\dfrac{{A(x)}}{{B(x)}}\] . It makes it easier to solve the integral as it is represented as the sum or difference of fractions. This method is only used when the fraction is linear in the denominator part.

Complete step by step answer:
The equation is as follows:
\[\int {\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)}}dx} \]
Now, we use partial fractions as,
\[\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)}} = \dfrac{A}{{2x - 1}} + \dfrac{B}{{x + 1}} + \dfrac{C}{{x - 1}}\]
Multiply both sides with \[\left( {2x - 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)\]
\[ \Leftrightarrow {x^2} + 1 = A\left( {x + 1} \right)\left( {x - 1} \right) + B\left( {2x - 1} \right)\left( {x - 1} \right) + C\left( {2x - 1} \right)\left( {x + 1} \right)\]
Expand the terms at the right side
\[ \Leftrightarrow {x^2} + 1 = A.{x^2} - A + B.2{x^2} - B.3x + B + C.2{x^2} + C.x - C\]

Collecting the terms we get as follows
\[ \Leftrightarrow {x^2} + 0.x + 1 = A{x^2} + 0.Ax - A + 2B{x^2} - 3Bx + B + 2C{x^2} + Cx - C\]
\[(I)1 = A + 2B + 2C..........{x^2}terms\]
\[(II)0 = - 3B + C...........xterms\]
\[(III)1 = - A + B - C\]
To solve this system of linear equations, you can e.g. transform (II) as \[C = 3B\]and \[3B\]for \[C\] into (I) and (III) obtaining as follows
\[1 = A + 8B\]
\[\Rightarrow 1 = - A - 2B\]

Adding those equation finally leads to \[B = \dfrac{1}{3}\] and thus, the solution of the linear equation system is:
\[A = - \dfrac{5}{3}\], \[B = \dfrac{1}{3}\] and \[C = 1\]
This means that your original fraction can be transformed as follows:
\[\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)}} = - \dfrac{5}{3}.\dfrac{1}{{2x - 1}} + \dfrac{1}{3}.\dfrac{1}{{x + 1}} + \dfrac{1}{{x - 1}}\]

So,the last thing to do is solving the integral
\[\int {\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)}}dx} \]
\[\Rightarrow \int {\left( { - \dfrac{5}{3}.\dfrac{1}{{2x - 1}} + \dfrac{1}{3}.\dfrac{1}{{x + 1}} + \dfrac{1}{{x - 1}}} \right)dx} \]
\[\Rightarrow - \dfrac{5}{3}\int {\dfrac{1}{{2x - 1}}dx + \dfrac{1}{3}\int {\dfrac{1}{{x + 1}}dx} + \int {\dfrac{1}{{x - 1}}dx} } \]
According to the formula \[\int {\dfrac{1}{x}dx} = \ln |x| + c\] we integrate the above as follows,
\[\therefore \dfrac{{ - 5}}{3}\ln |2x - 1|.\dfrac{1}{2} + \dfrac{1}{3}\ln |x + 1| + \ln |x - 1| + c\]

Hence, the integration of \[\int {\dfrac{{{x^2} + 1}}{{\left( {2x - 1} \right)\left( {x + 1} \right)\left( {x - 1} \right)}}dx} \] using partial fractions is $\dfrac{{ - 5}}{3}\ln |2x - 1|.\dfrac{1}{2} + \dfrac{1}{3}\ln |x + 1| + \ln |x - 1| + c$.

Note: If \[\dfrac{{A(x)}}{{B(x)}}\] is the equation then the partial fraction is applicable, if and only if, the degree of \[A(x)\] is less than that of \[B(x)\]. If the value is greater, then this method fails and we need to look for some other alternative. It is to be noted that integration and derivatives are the reverse of each other. If the derivative of a term a is b , then the integration of the term b will be a.And unlike derivatives, there is no chain rule to simplify the expression directly therefore, one of the simplification measures which is partial fraction is used here. Partial fraction is a method which involves expressing a fraction as a sum of two or more polynomials or fractions, for the purpose of simplification of the term to be integrated.