Question

How do you integrate $\dfrac{2}{{{x^3} + 1}}$ using partial fractions?

Answer 1

To integrate a given expression using partial fractions, firstly factorize the denominator of the integrand and use a partial fraction method to express it in two separate integrands whose denominators are the individual factors of the given integrand. And then integrate them separately.

Complete step by step solution:
In order to integrate $\dfrac{2}{{{x^3} + 1}}$ using partial fractions, we will first factorize the denominator of the given expression
$= {x^3} + 1$
Using algebraic identity ${a^3} + {b^3} = (a + b)\left( {{a^2} - ab + {b^2}} \right)$ to factorize it,
$= {x^3} + {1^3} \\\ = (x + 1)\left( {{x^2} - x + 1} \right) \\\$
Now, expressing the integrand as follows
$\dfrac{2}{{{x^3} + 1}} = \dfrac{2}{{(x + 1)\left( {{x^2} - x + 1} \right)}}$
Applying partial fraction decomposition, we will get
\dfrac{2}{{(x + 1)\left( {{x^2} - x + 1} \right)}} = \dfrac{2}{3}\left\\{ {\dfrac{1}{{x + 1}} - \dfrac{{x - 2}}{{\left( {{x^2} - x + 1} \right)}}} \right\\}
Taking integration both sides, we will get
\Rightarrow \int {\dfrac{2}{{(x + 1)\left( {{x^2} - x + 1} \right)}}dx} = \int {\dfrac{2}{3}\left\\{ {\dfrac{1}{{x + 1}} - \dfrac{{x - 2}}{{\left( {{x^2} - x + 1} \right)}}} \right\\}dx}
Distributing the integration above subtraction, we will get
$\Rightarrow \int {\dfrac{2}{{(x + 1)\left( {{x^2} - x + 1} \right)}}dx} = \dfrac{2}{3}\left[ {\int {\dfrac{1}{{x + 1}}dx} - \int {\dfrac{{x - 2}}{{\left( {{x^2} - x + 1} \right)}}dx} } \right]$
Now, factoring out the term where the numerator is a multiple of the derivative of the denominator
$\Rightarrow \int {\dfrac{2}{{(x + 1)\left( {{x^2} - x + 1} \right)}}dx} = \dfrac{2}{3}\left[ {\int {\dfrac{1}{{x + 1}}dx} - \dfrac{1}{2}\int {\dfrac{{x - 2}}{{\left( {{x^2} - x + 1} \right)}}dx} + \dfrac{3}{2}\int {\dfrac{1}{{\left( {{x^2} - x + 1} \right)}}dx} } \right]$
Completing the square at denominator, we will get
$\Rightarrow \int {\dfrac{2}{{(x + 1)\left( {{x^2} - x + 1} \right)}}dx} = \dfrac{2}{3}\left[ {\int {\dfrac{1}{{x + 1}}dx} - \dfrac{1}{2}\int {\dfrac{{x - 2}}{{\left( {{x^2} - x + 1} \right)}}dx} + \dfrac{3}{2}\int {\dfrac{4}{{{{\left( {2x - 1} \right)}^2} + 3}}dx} } \right]$
Integrating each term, we will get
$\Rightarrow \int {\dfrac{2}{{(x + 1)\left( {{x^2} - x + 1} \right)}}dx} = \dfrac{2}{3}\left[ {\ln \left| {x + 1} \right| - \dfrac{1}{2}\ln \left| {{x^2} - x + 1} \right| + \sqrt 3 {{\tan }^{ - 1}}\left( {\dfrac{{2x - 1}}{{\sqrt 3 }}} \right)} \right]$
Simplifying it,
$\Rightarrow \int {\dfrac{2}{{(x + 1)\left( {{x^2} - x + 1} \right)}}dx} = \dfrac{2}{3}\ln \left| {x + 1} \right| - \dfrac{1}{3}\ln \left| {{x^2} - x + 1} \right| + \dfrac{{2\sqrt 3 }}{3}{\tan ^{ - 1}}\left( {\dfrac{{2x - 1}}{{\sqrt 3 }}} \right)$

Therefore $\dfrac{2}{3}\ln \left| {x + 1} \right| - \dfrac{1}{3}\ln \left| {{x^2} - x + 1} \right| + \dfrac{{2\sqrt 3 }}{3}{\tan ^{ - 1}}\left( {\dfrac{{2x - 1}}{{\sqrt 3 }}} \right)$ is the integration of $\dfrac{{2x}}{{(x + 5)(x - 5)}}$

Additional Information:
Partial fractions method works only on proper rational fractions. Proper rational fraction is the fraction whose degree of the top (numerator part) should be less than that of the bottom part (denominator). And if you encounter a problem where the degree of top is more than the degree of bottom, then make the condition satisfied with help of division.

Note:
Take care of the signs when expressing the integrand in partial fraction form, and also cross check it once by adding or subtracting the partial fractions if the original integrand is obtained or not, because sometimes students calculate the incorrect partial fractions of the integrand and after integrating them get incorrect answers.