Definite integral (2x²−1)/(x²−3x+2) from −1 to 0

This problem asks to evaluate the definite integral of a rational function on the interval [−1, 0].

Problem

$\int_{- 1}^{0} \frac{2 x^{2} - 1}{x^{2} - 3 x + 2} d x$

Solution

Step 1

First note that the integrand is an improper rational fraction: the degree of the numerator (2) equals the degree of the denominator (2). Extract the polynomial part by dividing the numerator by the denominator:

Why do we extract the polynomial part?

A rational fraction P(x)/Q(x) is called proper if the degree of the numerator is less than the degree of the denominator, and improper otherwise. For integration, it is convenient to have a proper fraction — then it can be decomposed into partial fractions. If the fraction is improper, we first extract the polynomial part and turn the remainder into a proper fraction.

\frac{2 x^{2} - 1}{x^{2} - 3 x + 2} = 2 + \frac{6 x - 5}{x^{2} - 3 x + 2}

Step 2

The denominator factors as x² − 3x + 2 = (x − 2)(x − 1). Decompose the proper fraction into partial fractions:

How do we decompose a proper fraction into partial fractions?

Each linear factor (x − a) in the denominator corresponds to a partial fraction of the form A/(x − a). If a factor is repeated k times, it corresponds to k fractions: A₁/(x − a) + A₂/(x − a)² + … + Aₖ/(x − a)ᵏ. In our case, the denominator has two distinct linear factors — (x − 2) and (x − 1), so the decomposition has the form A/(x − 2) + B/(x − 1).

How do we find coefficients A and B?

The method of undetermined coefficients is used. Bring the sum of partial fractions to a common denominator and equate the numerators. We obtain the identity A(x−1) + B(x−2) = 6x − 5. Equating coefficients of equal powers of x gives a system of linear equations in A and B. Solving it, we find A = 7, B = −1.

\frac{6 x - 5}{(x - 2) (x - 1)} = \frac{A}{x - 2} + \frac{B}{x - 1}

Step 3

Expand the brackets and equate coefficients of equal powers of x. We obtain the system:

A (x - 1) + B (x - 2) = 6 x - 5

{\begin{matrix} A + B = 6 \\ - A - 2 B = - 5 \end{matrix}

Solving the system, we find: A = 7, B = −1.

Step 4

Substitute the decomposition into the integral:

\int_{- 1}^{0} (2 + \frac{7}{x - 2} - \frac{1}{x - 1}) d x

Step 5

Integrate each term and apply the Newton-Leibniz formula:

[_{- 1}^{0} 2 x + 7 ln | x - 2 | - ln | x - 1 |

Step 6

Substitute the limits of integration. We obtain the answer.

Answer

$2 - 7 ln 3 + 8 ln 2$

Frequently asked questions

How do I evaluate a definite integral of a rational function?

You need to: (1) if the fraction is improper — extract the polynomial part, (2) factor the denominator, (3) decompose the proper fraction into partial fractions, (4) find the coefficients by the method of undetermined coefficients, (5) integrate each term, (6) substitute the limits of integration using the Newton-Leibniz formula.

What is the method of undetermined coefficients?

It is a method of finding coefficients A, B, C… in the decomposition of a proper fraction into partial fractions. The idea: bring the sum of partial fractions to a common denominator, equate the numerators, and then equate coefficients of equal powers of x. This gives a system of linear equations in the unknown coefficients. An alternative is to substitute specific values of x (for example, the roots of the denominator) into the identity, which is often faster.

Why do logarithms appear in the answer?

Because after decomposition into partial fractions, we obtain integrals of the form ∫ A/(x − a) dx = A ln|x − a| + C. This is a standard situation when integrating rational functions with linear factors in the denominator.

What if the denominator has a repeated root?

If a linear factor (x − a) is repeated k times, it corresponds to a sum of k partial fractions: A₁/(x − a) + A₂/(x − a)² + … + Aₖ/(x − a)ᵏ. For example, for (x − 2)² the decomposition is A/(x − 2) + B/(x − 2)². This is shown in the next problem.