Integral Calculation Exercise 9

In this exercise, we aim to compute the following integral:

$$ \int \frac{x+2}{x^3 - x^2 - 2x} \, dx $$

We'll begin by decomposing the rational function into partial fractions.

The denominator is a cubic polynomial with distinct linear factors, each of multiplicity one.

Explanation. The polynomial has multiplicity 1 for each root, since each factor of the form $ (x - r) $ appears only once: $$ x^3 - x^2 - 2x = x \cdot (x^2 - x - 2) $$ Factoring the quadratic, we obtain: $$ x \cdot (x - 2)(x + 1) $$ The roots are $ r_1 = 0 $, $ r_2 = -1 $, and $ r_3 = 2 $. These are all simple roots (multiplicity 1).

We can now apply partial fraction decomposition as follows:

$$ \frac{x+2}{x^3 - x^2 - 2x} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 1} $$

where $ A $, $ B $, and $ C $ are constants to be determined.

To find these coefficients, we clear the denominator:

$$ \frac{x+2}{x^3 - x^2 - 2x} = \frac{A(x - 2)(x + 1) + Bx(x + 1) + Cx(x - 2)}{x(x - 2)(x + 1)} $$

Now we expand the numerator on the right-hand side:

$$ A(x^2 - x - 2) + B(x^2 + x) + C(x^2 - 2x) $$

Combining like terms yields:

$$ (A + B + C)x^2 + (B - A - 2C)x - 2A $$

Equating the numerators, we match coefficients term by term:

$$ \begin{cases} A + B + C = 0 \\ B - A - 2C = 1 \\ -2A = 2 \end{cases} $$

Explanation. Since the original numerator is $ x + 2 $, it has no $ x^2 $ term, which means the coefficient of $ x^2 $ must be zero. We apply the same reasoning for the coefficients of $ x $ and the constant term.

We solve the system using substitution. From the third equation, we find:

$$ A = -1 $$

Substitute into the first two equations:

$$ \begin{cases} -1 + B + C = 0 \Rightarrow B + C = 1 \\ B + 1 - 2C = 1 \Rightarrow B - 2C = 0 \end{cases} $$

From the second equation, we have $ B = 2C $. Substituting into the first:

$$ 2C + C = 1 \Rightarrow C = \frac{1}{3}, \quad B = \frac{2}{3} $$

So the constants are:

$$ A = -1, \quad B = \frac{2}{3}, \quad C = \frac{1}{3} $$

Substituting these into the original decomposition:

$$ \frac{x + 2}{x^3 - x^2 - 2x} = \frac{-1}{x} + \frac{2/3}{x - 2} + \frac{1/3}{x + 1} $$

We now rewrite the original integral as:

$$ \int \frac{x+2}{x^3 - x^2 - 2x} \, dx = \int \left( \frac{-1}{x} + \frac{2/3}{x - 2} + \frac{1/3}{x + 1} \right) dx $$

This allows us to break the expression into a sum of simpler integrals:

$$ -\int \frac{1}{x} \, dx + \frac{2}{3} \int \frac{1}{x - 2} \, dx + \frac{1}{3} \int \frac{1}{x + 1} \, dx $$

Each of these integrals is elementary:

$ \int \frac{1}{x} \, dx = \log |x| + c $
$ \int \frac{1}{x - 2} \, dx = \log |x - 2| + c $
$ \int \frac{1}{x + 1} \, dx = \log |x + 1| + c $

Putting everything together:

$$ \frac{2}{3} \log |x - 2| + \frac{1}{3} \log |x + 1| - \log |x| + c $$

This is the final result.

And so on.