Problem 34 - Calculus Early Transcendentals

Question

Find the following integrals. $$\int \frac{y^{2}}{(y+1)^{4}} d y$$

Step-by-Step Solution

Answer

Question: Find the integral of the given function: \(\int \frac{y^{2}}{(y+1)^{4}} dy\). Answer: \(-\frac{1}{y+1} + (y+1)^{2} -\frac{1}{3(y+1)^{3}} + C\)

Step 1: Perform substitution

Let \(u = y + 1\). Then, we have \(y = u - 1\). We also need the differential, so \(dy = du\). Now we can rewrite the integral in terms of \(u\):$$\int \frac{(u-1)^{2}}{u^{4}} du$$

Step 2: Expand the numerator

Expand the numerator of the fraction inside the integral:$$\int \frac{u^{2} - 2u + 1}{u^{4}} du$$

Step 3: Split the integral

Split the integral into three separate integrals:$$\int \frac{u^{2}}{u^{4}} du - 2\int \frac{u}{u^{4}} du + \int \frac{1}{u^{4}} du$$

Step 4: Simplify and integrate

Simplify the expressions inside the integrals and integrate each of them separately:$$\int u^{-2} du - 2\int u^{-3} du + \int u^{-4} du$$Now, integrate each of them:$$\int u^{-2}} du = \frac{u^{-1}}{-1} + C_{1} = -\frac{1}{u} + C_{1}$$$$ -2\int u^{-3} du = -\frac{u^{-2}}{-1} + C_{2} = u^{2} + C_{2}$$ $$\int u^{-4} du = \frac{u^{-3}}{-3} + C_{3} = -\frac{1}{3u^{3}} + C_{3}$$

Step 5: Combine the integrals

Combine the three integrals into one expression:$$-\frac{1}{u} + u^{2} -\frac{1}{3u^{3}} + C$$where \(C = C_{1} + C_{2} + C_{3}\) is the constant of integration.

Step 6: Substitute back for y

Substitute \(u = y + 1\) back into the expression to obtain the final result:$$-\frac{1}{y+1} + (y+1)^{2} -\frac{1}{3(y+1)^{3}} + C$$

Key Concepts

Integration by Substitution

Integration by substitution is a technique used in calculus to simplify the process of finding integrals. Think of it as changing variables to make the integration easier. For example, instead of integrating a complex function directly, you substitute a part of the function with a simpler variable. Here are the main steps involved:

• Identify a part of the integral to substitute with a new variable, say, \(u\).
• Express the original variable in terms of \(u\).
• Calculate the differential \(du\) in terms of the original differential \(dy\).
• Substitute all parts of the integral with their equivalents in terms of \(u\) and \(du\).
• Perform the integration with respect to \(u\).
• Finally, substitute back the expression for \(u\) in terms of the original variable to finish the integration.

This method leverages the chain rule in reverse, thus simplifying the integral calculation. In the problem given, substituting \(u = y + 1\) simplified the integral significantly, making it easier to handle.

Definite and Indefinite Integrals

Integrals are a fundamental concept in calculus, with two main types: definite and indefinite integrals.

An **indefinite integral** represents a family of functions and includes a constant of integration, \(C\). This is because the process of differentiation removes constants, so when integrating, you can have any constant added, and it would still satisfy the original function. The solution of the exercise is an indefinite integral as it involves the constant \(C\).

A **definite integral**, on the other hand, calculates the area under a curve between two points and therefore has limits of integration. It does not include a constant of integration. Definite integrals provide a numerical value rather than a function.

• Indefinite integrals: \(\int f(x) \, dx = F(x) + C\)
• Definite integrals: \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\)

Understanding the difference helps determine the approach and interpretation when solving integral problems.

Polynomial Long Division

Polynomial long division is a process similar to dividing numbers but applied to polynomials. It's useful for breaking down complex rational expressions into simpler parts, especially if you're dealing with integrals that have polynomial expressions in the numerator and denominator.

Here’s how it works:

• Divide the first term of the numerator by the first term of the denominator to get the first term of the quotient.
• Multiply the entire divisor by this term, and subtract the result from the original numerator to form a new polynomial.
• Repeat the process with the new polynomial as the numerator.

Each step reduces the degree of the polynomial until what's left is simpler to manage, often resulting in terms that can be more easily integrated or simplified further for the integral's solution. In the provided exercise, polynomial long division was bypassed because the substitution technique simplified the function, but this method was illustrated in breaking down fractions within the integrand.