Question

Evaluate the following integrals: $$ \int \frac{d x}{x\left[(\log x)^{2}+2 \log x-3\right]} $$

Short answer

Question: Evaluate the integral $$ \int \frac{dx}{x\left[(\log x)^{2}+2 \log x-3\right]} $$ Answer: $$ \int \frac{dx}{x\left[(\log x)^{2}+2 \log x-3\right]} = -\frac{1}{4}\log|\log x + 3| + \frac{1}{4}\log|\log x - 1| + C $$

Step-by-step solution

Factorize the quadratic equation

First, we need to factorize the quadratic equation inside the square brackets. Consider the expression: \((\log x)^2 + 2\log x - 3\). We can treat this as a quadratic equation in terms of \((\log x)\): $$ (\log x)^2 + 2\log x -3 = (\log x + 3)(\log x - 1) $$

Write the integral with the factorized quadratic equation

Now we have the factorized equation, let's rewrite the given integral as follows: $$ \int \frac{dx}{x\left[(\log x + 3)(\log x - 1)\right]} $$

Apply the substitution method

For this step, let's use the substitution method. Let \(u = \log x \Rightarrow x = e^u\). Now, let's find the derivative of this substitution with respect to \(x\): $$ \frac{du}{dx} = \frac{1}{x} \Rightarrow dx = x du $$ Now we rewrite the integral using the substitution \(u = \log x\): $$ \int \frac{dx}{x\left[(\log x + 3)(\log x - 1)\right]} = \int \frac{x du}{x\left[(u + 3)(u - 1)\right]} = \int \frac{du}{(u+3)(u-1)} $$

Apply partial fractions decomposition

Now, let's apply partial fractions decomposition to the expression in the integrand. We want to find constants \(A\) and \(B\) such that: $$ \frac{1}{(u + 3)(u - 1)} = \frac{A}{u + 3} + \frac{B}{u - 1} $$ Now, let's clear the fractions and solve for \(A\) and \(B\): $$ 1 = A(u - 1) + B(u + 3) $$ We can solve this equation by making \(u = 1\) and \(u = -3\). For \(u = 1\), we have \(A(1-1) + B(1+3) = 1\), which gives us \(B = \frac{1}{4}\). For \(u = -3\), we have \(A(-3-1) + B(-3+1) = 1\), which gives us \(A = -\frac{1}{4}\).

Write the integral after applying partial fractions

Now, let's rewrite the integral using the partial fractions we've found. The integral becomes: $$ \int \frac{du}{(u+3)(u-1)} = -\frac{1}{4}\int \frac{du}{u + 3} + \frac{1}{4}\int \frac{du}{u - 1} $$

Evaluate the integral

Now we can evaluate the integrals. As these are simple logarithmic integrals, their results are straightforward: $$ -\frac{1}{4}\int \frac{du}{u + 3} + \frac{1}{4}\int \frac{du}{u - 1} = -\frac{1}{4}\log|u + 3| + \frac{1}{4}\log|u - 1| + C $$ where \(C\) is the constant of integration.

Replace the substitution back

Finally, let's replace the substitution \(u = \log x\) back into the result to get the final result in terms of \(x\): $$ -\frac{1}{4}\log|u + 3| + \frac{1}{4}\log|u - 1| + C = -\frac{1}{4}\log|\log x + 3| + \frac{1}{4}\log|\log x - 1| + C $$ Thus, the final result for the integral is: $$ \int \frac{dx}{x\left[(\log x)^{2}+2 \log x-3\right]} = -\frac{1}{4}\log|\log x + 3| + \frac{1}{4}\log|\log x - 1| + C $$

Key concepts

Partial Fractions Decomposition

Partial fractions decomposition is a technique used in integral calculus to break down complex rational expressions into simpler fractions that are easier to integrate. This method is particularly useful when dealing with integrals of rational functions, where the numerator and the denominator are polynomials and the degree of the numerator is less than the degree of the denominator.

The process begins with factoring the denominator of the rational expression into its component parts, if possible. Once factored, we express the original fraction as a sum of fractions, each with one of these factors in its denominator. The numerators of these new fractions are typically unknown constants, which we determine by solving a system of equations generated from equating coefficients of like powers or by plugging in convenient values for the variables.

In the example provided, the denominator \( (\text{{log}} x)^2 + 2\text{{log}} x - 3 \) was factored as \( (\text{{log}} x + 3)(\text{{log}} x - 1) \) and then expressed as the sum of two simpler fractions. This step is crucial as it transforms the integral into a form that allowed us to apply further techniques for integration.

Substitution Method

The substitution method, often referred to as U-substitution, is a powerful tool in integral calculus that simplifies the integration process by changing the variable of integration to something more manageable. It closely resembles the chain rule for derivatives and is essentially its reverse application.

To apply the substitution method, we identify a part of the integrand that can be set equal to a new variable \( u \), and then express \( dx \) in terms of \( du \). This substitution should simplify the integral, making it easier to evaluate. In our specific example, the substitution \( u = \text{{log}} x \) was made. Consequently, \( dx \) is expressed as \( x du \) based on the derivative of \( u \) with respect to \( x \).

This substitution was a strategic move that simplified the integrand to a function of \( u \) alone, thus preparing the expression for the application of partial fractions decomposition and further integration.

Logarithmic Integration

Logarithmic integration is utilized when the integrand is in the form of a reciprocal function, such as \(1/x\) or \(1/u\), where \(u\) represents a linear function of the variable of integration. The result of integrating such a function is the natural logarithm of the absolute value of that variable.

In our exercise, after applying partial fractions decomposition and the substitution method, we obtained two integrals involving the reciprocal of linear terms in \( u \). Each integral of the form \( \text{{d}}u/(u + c) \) leads to \( \text{{log}}|u + c| \) up to a constant multiplier, where \( c \) is a constant. Here \( \text{{log}} \) denotes the natural logarithm, and the absolute value is necessary to cater for the situations where the argument inside the log might be negative.

The final solution required us to integrate the expressions \( -1/4\text{{d}}u/(u + 3) \) and \( 1/4\text{{d}}u/(u - 1) \) separately, yielding logarithmic functions of \( u \) that were then combined and rewritten in terms of the original variable \( x \) to arrive at the final integrated result.