Problem 56 - Calculus Single Variable

Question

Make a substitution before applying the method of partial fractions to calculate the given integral. $$ \int \frac{2 \ln (x)+5}{x\left(\ln ^{2}(x)+\ln (x)\right)} d x $$

Step-by-Step Solution

Answer

The integral equals \( 5 \ln |\ln(x)| - 3 \ln |\ln(x) + 1| + C \).

Step 1: Identify a Suitable Substitution

Notice that the integral contains a complex expression which includes \( \ln(x) \). To simplify the integral, let's set \( u = \ln(x) \). This means that when taking the derivative, \( du = \frac{1}{x} dx \), or equivalently, \( dx = x \, du \).

Step 2: Rewrite the Integral in Terms of the New Variable

Substitute \( u = \ln(x) \) and \( dx = x \, du \) into the integral. This transforms the integral to: \[ \int \frac{2u + 5}{x(u^2 + u)} x \, du = \int \frac{2u + 5}{u^2 + u} \, du \] by canceling the \( x \) terms.

Step 3: Simplify and Prepare for Partial Fractions

Rewrite the fraction \( \frac{2u + 5}{u^2 + u} \) so that it is ready to be decomposed into partial fractions. First, factor the denominator: \( u^2 + u = u(u+1) \).

Step 4: Perform Partial Fraction Decomposition

Express \( \frac{2u + 5}{u(u+1)} \) as a sum of partial fractions: \( \frac{A}{u} + \frac{B}{u+1} \). Set up the equation: \( 2u + 5 = A(u + 1) + Bu \).

Step 5: Solve for Coefficients A and B

Expand and collect like terms: \( 2u + 5 = Au + A + Bu \), thus \( 2u + 5 = (A + B)u + A \). From comparing coefficients, we have \( A + B = 2 \) and \( A = 5 \). Solve for \( A \) and \( B \):1. \( A = 5 \)2. Substitute \( A = 5 \) into \( A + B = 2 \) to find \( B = -3 \).

Step 6: Integrate the Partial Fractions

Substitute \( A = 5 \) and \( B = -3 \) back into the partial fractions: \( \int \left( \frac{5}{u} - \frac{3}{u+1} \right) \, du \). Integrate each term separately:- \( \int \frac{5}{u} \, du = 5 \ln |u| \)- \( -\int \frac{3}{u+1} \, du = -3 \ln |u+1| \).Combine results to get \( 5 \ln |u| - 3 \ln |u+1| + C \).

Step 7: Substitute Back to Original Variable

Since \( u = \ln(x) \), substitute back: \( 5 \ln |\ln(x)| - 3 \ln |\ln(x) + 1| + C \).

Key Concepts

Substitution Method

The substitution method is a technique used in calculus to simplify integrals by switching variables. This approach is particularly helpful when you encounter complex expressions or functions within the integral.

The core idea behind substitution is to transition from one variable to another, transforming the function into a simpler form. In the provided exercise, the integral comprises logarithmic components, suggesting that setting a new variable equal to the natural logarithm of the original variable can simplify the expression. Here, setting \( u = \ln(x) \) allows for a direct substitution that turns the integral into terms of \(u\). This eliminates the complexity of dealing with \(\ln(x)\) directly within the integral.

So to summarize, substitution involves:

• Choosing a suitable \(u\) that simplifies the original expression.
• Computing \( du \) in order to replace \( dx \) in the integral.
• Rewriting the integral completely in terms of the new variable \(u\).

The substitution method not only simplifies the process of integration but also lays the groundwork for applying further techniques, such as partial fraction decomposition.

Partial Fraction Decomposition

Partial fraction decomposition is a powerful tool used to break down rational expressions into simpler fractions. This technique is often applied after substitution has simplified the integral to a manageable form.

Once substitution has set up the integral, the resulting expression might still be complex due to polynomial denominators. In such cases, factoring the denominator into simpler components allows us to express the rational function as a sum of simpler fractions. This is exactly what happens if you rewrite \( \frac{2u + 5}{u^2 + u} \) into partial fractions.

When the denominator factors nicely, as in this case \( u(u+1) \), you can express the function as:

• \( \frac{A}{u} \) plus \( \frac{B}{u+1} \)

This allows further simplification as each component can be integrated separately. You find the coefficients \( A \) and \( B \) by equating the expression back to the original and solving the resulting equations. With \( A \) and \( B \) determined, the integral becomes easier to solve since each of these simpler fractions leads to recognizable basic integrals.

Logarithmic Integration

Logarithmic integration arises as a specific case of integrating functions which are derivatives of logarithmic functions. This technique becomes especially useful when you have already applied substitution and partial fraction decomposition.

In the given task, the partial fractions breakdown results in terms involving \( \frac{5}{u} \) and \( \frac{3}{u+1} \), both of which can be directly integrated using natural logarithm rules:

• The integral of \( \frac{5}{u} \) results in \( 5 \ln |u| \).
• The integral of \( \frac{3}{u+1} \) becomes \(-3 \ln |u+1| \).

This direct integration of these simple terms showcases logarithmic integration, rooted in the fundamental anti-derivative of \( \frac{1}{x} \), which is \( \ln |x| \).

After executing the logarithmic integration step, it’s crucial to substitute back to the original variable, replacing \(u\) with \(ln(x)\), completing the process and solving the integral.