Problem 43 - Calculus: One and Several Variables

Question

Derive the following three formulas. $$\int x^{k} \ln x d x=\frac{x^{k+1}}{k+1} \ln x-\frac{x^{k+1}}{(k+1)^{2}}+C, k \neq-1$$

Step-by-Step Solution

Answer

The derived formula is \( \int x^k \ln x dx = \frac{x^{k+1}}{k+1} \ln x - \frac{x^{k+1}}{(k+1)^2} + C \), for \( k \neq -1 \), which we found by applying integration by parts, choosing appropriate functions for u and dv, finding their derivatives and integrals, and simplifying the resulting expressions.

Step 1: Identify u and dv

Choose: \( u = \ln x \) and \( dv = x^k dx \) Now, we need to find du and v.

Step 2: Find du and v

Differentiate u with respect to x: \( du = \frac{1}{x} dx \) Integrate dv with respect to x: \( v = \int x^k dx = \frac{x^{k+1}}{k+1}, k\neq -1 \)

Step 3: Apply Integration by Parts

Use the integration by parts formula: \( \int x^k \ln x dx = uv - \int v du \) Substitute u, v, and du: \( \int x^k \ln x dx = \frac{x^{k+1}}{k+1} \ln x - \int \frac{x^{k+1}}{k+1} \cdot \frac{1}{x} dx \)

Step 4: Simplify the Integral

Simplify the second integral: \( \int x^k \ln x dx = \frac{x^{k+1}}{k+1} \ln x - \int \frac{x^{k}}{k+1} dx \)

Step 5: Integrate and Find the Constant of Integration

Integrate the second term: \( \int x^k \ln x dx = \frac{x^{k+1}}{k+1} \ln x - \frac{x^{k+1}}{(k+1)^2} + C \) The derived formula is given by: \( \int x^k \ln x dx = \frac{x^{k+1}}{k+1} \ln x - \frac{x^{k+1}}{(k+1)^2} + C \), for \( k \neq -1 \).

Key Concepts

Indefinite Integrals

Indefinite integrals are a fundamental concept in calculus, representing the antiderivative of a function. They are essentially the reverse operation of differentiation and are used to find the 'original' function given its derivative. Unlike definite integrals, which compute the area under a curve within certain limits, indefinite integrals do not have bounds and include a constant of integration, denoted as 'C', to account for any constant that was differentiated away.

The general symbol for an indefinite integral is \(\int f(x) dx\), which means to find all antiderivatives of the function \(f(x)\). The process of finding this integral involves several techniques, including basic antiderivatives, substitution, and, as in our example, integration by parts. The integration by parts is especially handy when dealing with the product of two functions, as in the given exercise.

Natural Logarithm Integration

Integrating functions that involve the natural logarithm, \(\ln x\), can initially seem daunting. However, there are strategies to make this process smoother. One of these strategies is recognizing when to apply the technique of integration by parts — particularly useful when the given function is a product of \(\ln x\) and a power of \(x\).

The natural logarithm function is the inverse of the exponential function \(e^x\) and its presence in an integral often signals the need for a more sophisticated approach than basic integration techniques. In our example problem, the integration by parts method elegantly handles the \(\ln x\) component by setting it as the 'u' in the formula \(\int u dv = uv - \int v du\). This method showcases how combining differential calculus (finding \(du\)) with integral calculus (finding \(v\)) creates a solution for integrating natural logarithms.

Calculus Formulas

Calculus is rich with formulas that help solve integrals and derivatives of functions. Some key formulas include the Power Rule for integration, \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), for \(n eq -1\), and the Exponential Function Rule, \(\int e^{ax}dx = \frac{1}{a}e^{ax} + C\). However, not all functions can be integrated using basic rules alone.

Integration by parts is itself a crucial formula derived from the product rule of differentiation, and it states \(\int u dv = uv - \int v du\). This formula allows us to break down more complex integrals into simpler parts and is particularly useful for integrals involving products of functions, such as polynomials and logarithms. Understanding and applying these calculus formulas is essential for tackling a wide range of problems, like the one provided, and mastering the subject of calculus as a whole.

Integration Techniques

There are various integration techniques to approach the multitude of functions one might encounter. Aside from basic antiderivatives and the integration by parts used in our example, other methods include:

• Substitution: Useful for functions that are the product of a function and its derivative, or to simplify the integral's form.
• Partial Fraction Decomposition: Applied to rational functions, where the denominator can be factored into linear or quadratic factors.
• Trigonometric Integration: For integrals involving trigonometric functions, sometimes requiring trigonometric identities for simplification.
• Trigonometric Substitution: In cases where the integrand includes \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\).

Each technique has its own place and utility. Developing the ability to recognize which technique to apply, as well as a familiarity with the various methods, is pivotal for students to successfully navigate through calculus problems. With practice, students will learn to identify patterns and select the most efficient method to find the integral.