Integral Calculation Exercise 42
We're asked to evaluate the integral
$$ \int \frac{x^2-3}{x^3-4x^2+5x-2} \ dx $$
The integrand is a rational function where the degree of the denominator exceeds that of the numerator, so polynomial division isn’t applicable here.
To compute this integral, we'll apply the method of partial fraction decomposition.
The denominator vanishes at \( x = 1 \), which we can exploit to factor the polynomial using Ruffini's rule.
$$ \begin{array}{c|lcc|r} & 1 & -4 & 5 & -2 \\ 1 & & 1 & -3 & 2 \\ \hline & 1 & -3 & 2 & 0 \end{array} $$
So the denominator factors as (x-1)(x2-3x+2):
$$ x^3 - 4x^2 + 5x - 2 = (x - 1)(x^2 - 3x + 2) $$
The quadratic factor \( x^2 - 3x + 2 \) has a positive discriminant, so it admits two distinct real roots:
$$ \Delta = b^2 - 4ac = (-3)^2 - 4(1)(2) = 9 - 8 = 1 $$
Thus, the solutions are \( x_1 = 1 \) and \( x_2 = 2 \):
$$ x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{3 \pm 1}{2} = \begin{cases} x_1 = 1 \\ \\ x_2 = 2 \end{cases} $$
We can now write the factorization as (x - 1)(x - 2), and hence:
$$ x^3 - 4x^2 + 5x - 2 = (x - 1)^2 (x - 2) $$
Substituting this into the original integral gives:
$$ \int \frac{x^2 - 3}{x^3 - 4x^2 + 5x - 2} \ dx = \int \frac{x^2 - 3}{(x - 1)^2 (x - 2)} \ dx $$
This setup is ideal for applying partial fraction decomposition:
$$ \int \frac{x^2 - 3}{(x - 1)^2 (x - 2)} \ dx = \int \left( \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x - 2} \right) dx $$
We determine the constants A, B, and C by expressing the integrand as a single rational function:
$$ \frac{x^2 - 3}{(x - 1)^2 (x - 2)} = \frac{A(x - 1)(x - 2) + B(x - 2) + C(x - 1)^2}{(x - 1)^2 (x - 2)} $$
Expanding the numerator:
$$ A(x^2 - 3x + 2) + B(x - 2) + C(x^2 - 2x + 1) $$
Combining like terms:
$$ (A + C)x^2 + (-3A + B - 2C)x + (2A - 2B + C) $$
Matching coefficients with the original numerator \( x^2 - 3 \), we obtain the system:
$$ \begin{cases} A + C = 1 \\ -3A + B - 2C = 0 \\ 2A - 2B + C = -3 \end{cases} $$
Solving this system step by step yields:
$$ \begin{cases} A = 1 - C \\ C = 3 - B \\ B = 2 \end{cases} \Rightarrow \begin{cases} A = 0 \\ C = 1 \\ B = 2 \end{cases} $$
Substituting the values back into the decomposition gives:
$$ \int \frac{x^2 - 3}{(x - 1)^2 (x - 2)} \ dx = \int \left( \frac{2}{(x - 1)^2} + \frac{1}{x - 2} \right) dx $$
Using the linearity property of integrals, we split the integral:
$$ \int \frac{x^2 - 3}{(x - 1)^2 (x - 2)} \ dx = 2 \int \frac{1}{(x - 1)^2} \ dx + \int \frac{1}{x - 2} \ dx $$
Both integrals are standard:
- The first is a power integral: $$ \int (x - 1)^{-2} \ dx = -\frac{1}{x - 1} + C $$
- The second is a logarithmic integral: $$ \int \frac{1}{x - 2} \ dx = \log |x - 2| + C $$
Putting everything together:
$$ \int \frac{x^2 - 3}{(x - 1)^2 (x - 2)} \ dx = -\frac{2}{x - 1} + \log |x - 2| + C $$
This is the final result of the integral.
And so on...
