Examples, Detailed Solutions, and Calculus Exercises
Calculate integrals of rational functions using partial fractions decomposition. This tutorial presents examples with detailed solutions, including instances where the degree of the numerator is greater than or equal to the degree of the denominator.
Note: In what follows, \( C \) represents the constant of integration.
You can also use an online partial fractions decomposition calculator to help decompose rational functions.
We use partial fractions decomposition to decompose the integrand into simpler fractions:
\[ \dfrac{-5x + 11}{x^2+x-2} = \dfrac{2}{x-1} - \dfrac{7}{x+2} \]We now use a table of integrals to integrate:
\[ \int \dfrac{-5x + 11}{x^2+x-2} dx = \int \dfrac{2}{x-1} dx - \int \dfrac{7}{x+2} dx \] \[ = 2 \ln|x - 1| - 7 \ln|x+2| + C \]A partial fractions decomposition of the integrand gives:
\[ \dfrac{x^2+6x - 3}{(x+3)(x^2+2x+9)} = \dfrac{2x+2}{x^2+2x+9} - \dfrac{1}{x+3} \]We now use a table of integrals to evaluate the integrals:
\[ \displaystyle \int \dfrac{x^2+6x - 3}{(x+3)(x^2+2x+9)} dx = \int \dfrac{2x+2}{x^2+2x+9} dx - \int \dfrac{1}{x+3} dx \] \[ = \ln|x^2+2x+9| - \ln|x+3| + C \]In this example, the degree of the numerator is greater than the degree of the denominator, and therefore a division of the numerator by the denominator is carried out in order to write the integrand as follows:
\[ \dfrac{2x^3 + 10x^2 +11x}{x^2+5x+6} = 2x - \dfrac{x}{x^2+5x+6} \]A partial fractions decomposition of the term \( \dfrac{x}{x^2+5x+6} \) gives:
\[ \dfrac{2x^3 + 10x^2 +11x}{x^2+5x+6} = 2x - \dfrac{x}{x^2+5x+6} = 2x +\dfrac{2}{x+2}-\dfrac{3}{x+3} \]Using the above, the given integral may be written as:
\[ \displaystyle \int \dfrac{2x^3 + 10x^2 +11x}{x^2+5x+6} dx = \int 2x dx +\int \dfrac{2}{x+2} dx - \int \dfrac{3}{x+3} dx \]Using a table of integrals, we evaluate the integrals as follows:
\[ = x^2 + 2 \ln|x+2| - 3 \ln|x+3| + C \]Evaluate the following integrals using partial fractions. Expand the boxes to reveal the step-by-step solutions.
Decompose into simpler fractions:
\[ \dfrac{-x + 7}{x^2+x-2} = \dfrac{2}{x-1} - \dfrac{3}{x+2} \]Hence,
\[ \int \dfrac{-x + 7}{x^2+x-2} dx = \int \left(\dfrac{2}{x-1} - \dfrac{3}{x+2}\right) dx \] \[ = 2 \ln|x-1| - 3 \ln|x+2| + C \]Decompose into simpler fractions:
\[ \dfrac{-8x^2 +23x - 5}{(x+7)(2x^2+x+2)} = -\dfrac{6}{x+7} + \dfrac{4x+1}{2x^2+x+2} \]Hence,
\[ \int \dfrac{-8x^2 +23x - 5}{(x+7)(2x^2+x+2)} dx = \int \left(-\dfrac{6}{x+7} + \dfrac{4x+1}{2x^2+x+2}\right) dx \] \[ = \ln|2x^2+x+2| - 6 \ln|x+7| + C \]Because the degree of the numerator is higher, divide first, then decompose into simpler fractions:
\[ \dfrac{x^4+3x^3+2x^2+7x+9}{x^2+3x+2} = x^2 + \dfrac{2}{x+1} + \dfrac{5}{x+2} \]Hence,
\[ \displaystyle \int \dfrac{x^4+3x^3+2x^2+7x+9}{x^2+3x+2} dx = \int \left(x^2 + \dfrac{2}{x+1} + \dfrac{5}{x+2}\right) dx \] \[ = \dfrac{x^3}{3}+2\ln |x+1|+5 \ln |x+2| + C \]