Question

How do you evaluate the integral $\int {\dfrac{1}{{{x^3} - x}}dx} $?

Answer 1

Hint: This question will be solved by using the partial fraction method. To apply this method, we have to convert the denominator into its simplest form. Then decompose the integrand and simplify it. After that, take different values of x and find A, B, and C. Substitute the decomposition into the integral. Then apply the formula as stated below.
$\int {\dfrac{1}{x}dx = \ln \left| x \right|} $

Complete step-by-step answer:
Let us use a partial fraction method to solve this question.
$ \Rightarrow \int {\dfrac{1}{{{x^3} - x}}dx} $
Now, take out a common factor x from the denominator.
$ \Rightarrow \int {\dfrac{1}{{x\left( {{x^2} - 1} \right)}}dx} $
As we already know the algebraic formula$\left( {{x^2} - 1} \right) = \left( {x - 1} \right)\left( {x + 1} \right)$.
Let us substitute that formula at the denominator.
$ \Rightarrow \int {\dfrac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}dx} $
Let us decompose the integrand.
$ \Rightarrow \dfrac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{A}{x} + \dfrac{B}{{x - 1}} + \dfrac{C}{{x + 1}}$
Here, we will do the lowest common denominator of all the terms.
$ \Rightarrow \dfrac{1}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{{A\left( {x - 1} \right)\left( {x + 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} + \dfrac{{B\left( x \right)\left( {x + 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} + \dfrac{{C\left( x \right)\left( {x - 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}$
Now, we can remove all terms which have an equal denominator.
$ \Rightarrow 1 = A\left( {x - 1} \right)\left( {x + 1} \right) + B\left( x \right)\left( {x + 1} \right) + C\left( x \right)\left( {x - 1} \right)$ ...(1)
Let us find the values of A, B, and C.
For that, we will take different values of x.
Let us take $x = 0$ in equation (1).
$ \Rightarrow 1 = A\left( {0 - 1} \right)\left( {0 + 1} \right) + B\left( 0 \right)\left( {0 + 1} \right) + C\left( 0 \right)\left( {0 - 1} \right)$
Therefore,
$ \Rightarrow 1 = A\left( { - 1} \right)\left( 1 \right)$
Finally,
$ \Rightarrow A = - 1$
Let us take $x = 1$ in equation (1).
$ \Rightarrow 1 = A\left( {1 - 1} \right)\left( {1 + 1} \right) + B\left( 1 \right)\left( {1 + 1} \right) + C\left( 1 \right)\left( {1 - 1} \right)$
Therefore,
$ \Rightarrow 1 = B\left( 1 \right)\left( 2 \right)$
$ \Rightarrow 1 = 2B$
Now, we will divide both sides by 2.
Finally,
$ \Rightarrow B = \dfrac{1}{2}$
Let us take $x = - 1$ in equation (1).
$ \Rightarrow 1 = A\left( { - 1 - 1} \right)\left( { - 1 + 1} \right) + B\left( { - 1} \right)\left( { - 1 + 1} \right) + C\left( { - 1} \right)\left( { - 1 - 1} \right)$
Therefore,
$ \Rightarrow 1 = C\left( { - 1} \right)\left( { - 2} \right)$
$ \Rightarrow 1 = 2C$
Now, we will divide both sides by 2.
Finally,
$ \Rightarrow C = \dfrac{1}{2}$
Let us put the decomposition into the integral.
$ \Rightarrow \int {\left( {\dfrac{{ - 1}}{x} + \dfrac{{\dfrac{1}{2}}}{{x - 1}} + \dfrac{{\dfrac{1}{2}}}{{x + 1}}} \right)dx} $
Now, apply integration to each term.
$ \Rightarrow \int {\dfrac{{ - 1}}{x}dx} + \int {\dfrac{{\dfrac{1}{2}}}{{x - 1}}dx + } \int {\dfrac{{\dfrac{1}{2}}}{{x + 1}}dx} $
Take out constant terms from the integral.
$ \Rightarrow \left( { - 1} \right)\int {\dfrac{1}{x}dx} + \dfrac{1}{2}\int {\dfrac{1}{{x - 1}}dx + } \dfrac{1}{2}\int {\dfrac{1}{{x + 1}}dx} $
Now, we will apply the integral formula as stated below:
$\int {\dfrac{1}{x}dx = \ln \left| x \right|} $
So, we get the answer as below.

$ \Rightarrow - \ln \left| x \right| + \dfrac{1}{2}\ln \left| {x - 1} \right| + \dfrac{1}{2}\ln \left| {x + 1} \right| + C$

Note:
In a partial fraction, the numerator must be of at least one degree less than the denominator. The denominator must contain linear factors, repeated linear factors, or a quadratic factor.
We decompose the fractions into partial fractions because:
> It makes certain integrals much easier to do.
> It is used in the Laplace transform.