Integral Calculation Exercise 16

We are asked to evaluate the following integral:

$$ \int \frac{x - 2}{x^3 - 7x - 6} \, dx $$

To begin, we factor the denominator using Ruffini’s synthetic division.

Since the polynomial $ x^3 - 7x - 6 $ has a root at $ x = 3 $, it is divisible by $ (x - 3) $.

Performing the division:

$$ \begin{array}{c|lcc|r} & 1 & 0 & -7 & -6 \\ 3 & & 3 & 9 & 6 \\ \hline & 1 & 3 & 2 & 0 \end{array} $$

This gives us the factorization:

$$ x^3 - 7x - 6 = (x - 3)(x^2 + 3x + 2) $$

Next, we factor the quadratic:

$$ x^2 + 3x + 2 = (x + 1)(x + 2) $$

Substituting into the integral, we obtain:

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

We now apply partial fraction decomposition:

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

Multiplying through by the common denominator:

$$ x - 2 = A(x + 2)(x + 1) + B(x - 3)(x + 1) + C(x - 3)(x + 2) $$

Expanding each term:

$$ A(x^2 + 3x + 2) + B(x^2 - 2x - 3) + C(x^2 - x - 6) $$

Combining like terms:

$$ (A + B + C)x^2 + (3A - 2B - C)x + (2A - 3B - 6C) $$

We now compare coefficients with the original numerator $ x - 2 $. This gives us the system:

Solving the system using substitution: From the first equation, we have $ A = -B - C $.

Substitute into the other two equations:

$$ 3(-B - C) - 2B - C = 1 \quad \Rightarrow \quad -5B - 4C = 1 $$

$$ 2(-B - C) - 3B - 6C = -2 \quad \Rightarrow \quad -5B - 8C = -2 $$

Subtracting the first from the second:

$$ (-5B - 8C) - (-5B - 4C) = -2 - 1 \quad \Rightarrow \quad -4C = -3 \quad \Rightarrow \quad C = \frac{3}{4} $$

Substitute back to find $ B $:

$$ -5B - 4 \cdot \frac{3}{4} = 1 \quad \Rightarrow \quad -5B - 3 = 1 \quad \Rightarrow \quad B = -\frac{4}{5} $$

Now solve for $ A $:

$$ A = -B - C = -(-\frac{4}{5}) - \frac{3}{4} = \frac{4}{5} - \frac{3}{4} = \frac{1}{20} $$

We now rewrite the integral as:

$$ \int \frac{1}{20(x - 3)} - \frac{4}{5(x + 2)} + \frac{3}{4(x + 1)} \, dx $$

By linearity, this becomes:

$$ \frac{1}{20} \int \frac{1}{x - 3} \, dx - \frac{4}{5} \int \frac{1}{x + 2} \, dx + \frac{3}{4} \int \frac{1}{x + 1} \, dx $$

Each integral is elementary:

Therefore, the final result is:

$$ \int \frac{x - 2}{x^3 - 7x - 6} \, dx = \frac{1}{20} \log|x - 3| - \frac{4}{5} \log|x + 2| + \frac{3}{4} \log|x + 1| + c $$

And so on.