Problem 54 - Calculus Single Variable

Question

Calculate the integrals. $$ \int \frac{x^{2}}{(x-1)^{3}} d x $$

Step-by-Step Solution

Answer

\( \ln|x-1| - \frac{2}{x-1} - \frac{1}{2(x-1)^2} + C \)

Step 1: Prepare for Substitution

To handle the integral \( \int \frac{x^{2}}{(x-1)^{3}} \, dx \), consider a substitution to simplify the denominator \((x-1)\). Let \( u = x - 1 \), which gives \( du = dx \) and \( x = u + 1 \). This substitution simplifies the original integral.

Step 2: Substitute into the Integral

Substitute \( x = u + 1 \) and \( dx = du \) into the integral, giving \( \int \frac{(u + 1)^{2}}{u^{3}} \, du \). Expand the numerator to \((u^2 + 2u + 1)\). Thus, the integral becomes \( \int \frac{u^2 + 2u + 1}{u^3} \, du \).

Step 3: Simplify the Expression

Break down the integral \( \int \frac{u^2}{u^3} + \frac{2u}{u^3} + \frac{1}{u^3} \, du \) into separate integrals: \( \int u^{-1} \, du + 2 \int u^{-2} \, du + \int u^{-3} \, du \).

Step 4: Integrate Each Term

Evaluate each integral separately: \( \int u^{-1} \, du = \ln|u| \), \( 2 \int u^{-2} \, du = -2u^{-1} \), and \( \int u^{-3} \, du = -\frac{1}{2}u^{-2} \). So, the combined solution is \( \ln|u| - 2u^{-1} - \frac{1}{2}u^{-2} + C \).

Step 5: Substitute Back to Original Variable

Replace \( u \) with \( x-1 \) to express the solution in terms of \( x \): \( \ln|x-1| - \frac{2}{x-1} - \frac{1}{2(x-1)^2} + C \). This is the final answer to the integral.

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus, especially for integration. It involves replacing a part of the equation with a new variable to simplify the integration process. This method is particularly helpful when dealing with complex expressions or where the integration of a function seems challenging.Here's how it generally works:

• Identify a Substitution: Choose a part of the integrand, usually within a composite function or the denominator in rational functions, and let it equal a new variable.
• Differentiate Substitution: Find the derivative of the substitution to express the differential in terms of the new variable.
• Replace and Simplify: Substitute both the original variable and the differential, adjusting the integrand as needed. Simplify the resulting integral.

Using this method allows you to convert a difficult integral into a simpler one, making the integration process more straightforward. In our specific problem, substituting \( u = x - 1 \) into \( \int \frac{x^{2}}{(x-1)^{3}} \, dx \) turns it into a simpler expression, which is easier to manage. Remember that choosing the right substitution is key to solving integrals efficiently.

Rational Functions

Rational functions are fractions involving polynomials in the numerator and the denominator. Understanding how to manipulate and integrate these functions is crucial in calculus, as they frequently appear in various problems.Some key points about rational functions include:

• Form of a Rational Function: Typically presented as \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials.
• Simplification: Often, simplifying the expression or performing long division can make integrating these functions more manageable.
• Usage of Partial Fractions: This technique is sometimes used to decompose a complex rational function into simpler fractions that are easier to integrate separately.

In our exercise, the rational function is \( \frac{x^{2}}{(x-1)^{3}} \). By making a smart substitution like \( u = x - 1 \), it simplifies the expression, turning it into \( \int \frac{u^2 + 2u + 1}{u^3} \, du \). This step then allows us to separate it into easier integrals, facilitating a straightforward path to the solution.

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, represent a family of functions that, when differentiated, yield the original function. Finding these integrals is one of the core tasks in calculus, revealing a function's accumulation pattern without specific limits.Here's what you need to know about indefinite integrals:

• General Form: Denoted by \( \int f(x) \, dx \), represents all possible antiderivatives of \( f(x) \).
• Constant of Integration: The result includes an arbitrary constant \( C \), reflecting all possible vertical shifts of the function.
• Techniques: Numerous methods exist to solve them, such as substitution, integration by parts, and special formulae for trigonometric, exponential, and logarithmic functions.

In the provided example, once a suitable substitution was made, we integrated the expression \( \frac{u^2 + 2u + 1}{u^3} \) separately as simpler fractions, resulting in \( \ln|u| - 2u^{-1} - \frac{1}{2}u^{-2} + C \). Finally, substituting back \( u = x - 1 \) produced the indefinite integral in terms of \( x \), demonstrating the transformation and integration of the function step by step.