Definite integral x/(x²−4x+4) from 3 to 4

This problem asks to evaluate the definite integral of a rational function with a repeated root of the denominator.

Problem

$\int_{3}^{4} \frac{x}{x^{2} - 4 x + 4} d x$

Solution

Step 1

The denominator is a perfect square:

Why does the denominator have a repeated root?

The discriminant of the quadratic x² − 4x + 4 is zero: D = 16 − 16 = 0. This means the quadratic has a single (double) root x = 2 and factors as (x − 2)². In this case, we say the root is repeated, and the partial fraction decomposition has its own peculiarities.

x^{2} - 4 x + 4 = {(x - 2)}^{2}

Step 2

Since the denominator has a repeated root, the partial fraction decomposition has the form:

How do we decompose into partial fractions with a repeated root?

If a linear factor (x − a) is repeated k times, it corresponds to a sum of k partial fractions with that denominator in different powers: A₁/(x − a) + A₂/(x − a)² + … + Aₖ/(x − a)ᵏ. In our case, the denominator is (x − 2)², so the decomposition is A/(x − 2) + B/(x − 2)².

\frac{x}{{(x - 2)}^{2}} = \frac{A}{x - 2} + \frac{B}{{(x - 2)}^{2}}

Step 3

Bring to a common denominator and equate the numerators:

A (x - 2) + B = x

From this we obtain A = 1, B = 2.

Step 4

Substitute the decomposition into the integral:

\int_{3}^{4} (\frac{1}{x - 2} + \frac{2}{{(x - 2)}^{2}}) d x

Step 5

Integrate: ∫ 1/(x − 2) dx = ln|x − 2|, ∫ 2/(x − 2)² dx = −2/(x − 2). Apply the Newton-Leibniz formula:

[_{3}^{4} ln | x - 2 | - \frac{2}{x - 2}

Step 6

Substitute the limits of integration. We obtain the answer.

Answer

$1 + ln 2$

Frequently asked questions

What is a repeated root of the denominator?

This is when the quadratic trinomial (or its factor) has a discriminant equal to zero, and can be written as (x − a)². In this case, the root x = a is called a double root (or root of multiplicity 2). In the partial fraction decomposition, a repeated root requires not one but several fractions with increasing powers.

Why do two terms appear in the decomposition?

Because the denominator (x − 2)² is a repeated root. Each factor in the denominator gives a partial fraction, and if the factor is repeated k times, it corresponds to k partial fractions with denominators (x − 2), (x − 2)², …, (x − 2)ᵏ. In our case k = 2.

How do we find coefficients A and B?

By the method of undetermined coefficients. Equate the numerators after bringing to a common denominator: A(x − 2) + B = x. Equating coefficients of x and free terms: A = 1, −2A + B = 0, hence A = 1, B = 2. Alternatively, substitute x = 2 (root of the denominator) — we get B = 2, then x = 0 — we get −2A + 2 = 0, A = 1.

Why does the integral of 2/(x − 2)² not contain a logarithm?

Because ∫ (x − 2)⁻² dx = −(x − 2)⁻¹ = −1/(x − 2). The power of x − 2 is not equal to −1, so the formula ∫ uⁿ du = uⁿ⁺¹/(n + 1) gives a rational result without a logarithm. A logarithm appears only when n = −1.