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

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

Problem

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

Solution

Step 1

The denominator is a perfect square:

Why does the denominator have a repeated root?

The discriminant of the quadratic x² + 2x + 1 is zero: D = 4 − 4 = 0. This means the quadratic has a single (double) root x = −1 and factors as (x + 1)². In this case, we say the root is repeated, and the partial fraction decomposition has its own peculiarities.

{(x + 1)}^{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 + 1)², so the decomposition is A/(x + 1) + B/(x + 1)².

\frac{x}{{(x + 1)}^{2}} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}}

Step 3

Bring to a common denominator and equate the numerators:

A (x + 1) + B = x

From this we obtain A = 1, B = −1.

Step 4

Substitute the decomposition into the integral:

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

Step 5

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

[_{0}^{1} ln | x + 1 | + \frac{1}{x + 1}

Step 6

Substitute the limits. ln 1 = 0, so we get ln 2 + 1/2 − 1 = ln 2 − 1/2.

Answer

$ln 2 - \frac{1}{2}$

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 + 1)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 + 1), (x + 1)², …, (x + 1)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 + 1) + B = x. Equating coefficients of x and free terms: A = 1, A + B = 0, hence A = 1, B = −1. Alternatively, substitute x = −1 (root of the denominator) — we get B = −1, then x = 0 — we get A + B = 0, A = 1.

Why does the integral of 1/(x+1)² not contain a logarithm?

Because ∫ (x + 1)⁻² dx = −(x + 1)⁻¹ = −1/(x + 1). The power of x + 1 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.