Find the partial fractions decomposition and an antiderivative. \frac{x + 2}{x^3 + x}

Question:

Find the partial fractions decomposition and an antiderivative.

{eq}\frac{x + 2}{x^3 + x} {/eq}

Rational Functions:

Rational fractions, quotients of polynomials, are broken down into partial fractions.

With this type of decomposition we can solve the indefinite integrals of this type of functions in a very simple way.

First, we need to decompose the denominator in irreductible factors:

{eq}\frac{{x + 2}}{{{x^3} + x}} = \frac{{x + 2}}{{x\left( {{x^2} + 1} \right)}} {/eq}

Using partial fractions and equating numerators:

{eq}\frac{{x + 2}}{{x\left( {{x^2} + 1} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}}\\ \frac{{x + 2}}{{x\left( {{x^2} + 1} \right)}} = \frac{{A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)x}}{{x\left( {{x^2} + 1} \right)}}\\ \\ x + 2 = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)x {/eq}

Calculating the constants, we obtain the decomposition:

{eq}x + 2 = A{x^2} + A + B{x^2} + Cx\\ \\ \left\{ \begin{array}{l} 0 = A + B\\ 1 = C\\ 2 = A \end{array} \right.\\ \\ A = 2,B = - 2,C = 1\\ \\ \frac{{x + 2}}{{x\left( {{x^2} + 1} \right)}} = \frac{2}{x} + \frac{{ - 2x + 1}}{{{x^2} + 1}} {/eq}