Solving the Integral ∫ x / (2x^2 - x - 1) dx: A Step-by-Step Guide

In this article, we will walk through the process of evaluating the integral ∫ x / (2x^2 - x - 1) dx. We'll start by factoring the denominator and then use partial fraction decomposition to simplify the integrand. Finally, we will integrate the simplified terms. Let's dive into the details!

1. Factoring the Denominator

The denominator of the integrand is a quadratic expression: 2x^2 - x - 1. The first step is to factor this quadratic into two simpler linear factors. We need to find two numbers that multiply to 2 × -1 -2 and add to -1. These numbers are -2 and 1.

Thus, we can rewrite the quadratic as:

2x^2 - x - 1  2x^2 - 2x   x - 1  2x(x - 1)   (x - 1)  (2x   1)(x - 1)

2. Partial Fraction Decomposition

Next, we can express the integrand using partial fractions:

int frac{x}{2x  1x - 1} dx  int left[ frac{A}{2x   1}   frac{B}{x - 1} right] dx

Multiplying through by the denominator (2x 1)(x - 1), we get:

x  A(x - 1)   B(2x   1)

Expanding and simplifying, we obtain:

x  Ax - A   2Bx   B  (A   2B)x   (B - A)

Equating coefficients, we have:

From B - A 0, we find B A. Substituting B A into the first equation gives:

A   2A  1 implies 3A  1 implies A  1/3

Therefore, we have A 1/3 and B 1/3. So, the integrand can be expressed as:

frac{x}{(2x   1)(x - 1)}  frac{1/3}{2x   1}   frac{1/3}{x - 1}

3. Integration

Now we can integrate each term separately:

int frac{x}{2x   1x - 1} dx  int left( frac{1/3}{2x   1}   frac{1/3}{x - 1} right) dx

This gives:

1/3 int frac{1}{2x   1} dx   1/3 int frac{1}{x - 1} dx

The integrals evaluate as follows:

We use the substitution u 2x 1, hence du 2 dx or dx frac{du}{2}:

int frac{1}{2x   1} dx  frac{1}{2} int frac{1}{u} du  frac{1}{2} ln|2x   1|   C_1

The integral is:

int frac{1}{x - 1} dx  ln|x - 1|   C_2

Putting everything together, we have:

int frac{x}{2x^2 - x - 1} dx  frac{1}{3} left( frac{1}{2} ln|2x   1|   ln|x - 1| right)   C

This simplifies to:

1/6 ln|2x   1|   1/3 ln|x - 1|   C

4. Final Result

Thus, the solution to the integral is:

int frac{x}{2x^2 - x - 1} dx  1/6 ln|2x   1|   1/3 ln|x - 1|   C

This is the final result for the integral we were evaluating. The final answer is expressed in terms of natural logarithms and a constant of integration C.

Integrals like this demonstrate the power of partial fraction decomposition and the integration by substitution technique. Understanding these concepts is crucial for solving more complex integrals.