Problem 41 - Calculus: One and Several Variables

Question

Derive the formula.$$\int \frac{d u}{u^{2}(a+b u)}=-\frac{1}{a u}+\frac{b}{a^{2}} \ln \left|\frac{a+b u}{u}\right|+C$$.

Step-by-Step Solution

Answer

To derive the formula, we perform integration by partial fractions. First, we rewrite the integrand as \(-\frac{1}{au} + \frac{b}{a^2(a+bu)}\). Then, we integrate term by term and simplify the expression by combining the logarithms, resulting in the final expression: \(-\frac{1}{au}+\frac{b}{a^{2}} \ln \left|\frac{a+b u}{u}\right|+C\).

Step 1: Rewrite the integrand as partial fractions

We are given the integrand, \(\frac{1}{u^2(a+bu)}\). Our goal is to rewrite it as a sum of simpler fractions, so we can easily integrate term by term. We can express it as the following: $$\frac{1}{u^2(a+bu)} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{a+bu}$$ Now, we need to determine the values of A, B, and C. Step 2: Determine the coefficients

Step 2: Solve for the coefficients A, B, and C

To find the coefficients A, B, and C, we will multiply both sides by the common denominator, \(u^2(a+bu)\): $$1 = A(u^2)(a+bu) + B(a+bu)u + Cu^2$$ Next, we will set up three equations to solve for A, B, and C: 1. Let u = 0, then 1 = A(a(0)) Thus, \(A = -\frac{1}{a}\) 2. Differentiate both sides of the equation with respect to u and let u = 0, then 0 = B(a). Thus, \(B = 0\) 3. Differentiate both sides of the equation with respect to u once more and let u = 0, then 0 = 2C. Thus, \(C = \frac{b}{a^2}\) Now that we have found the coefficients, we can rewrite our partial fraction decomposition as: $$\frac{1}{u^2(a+bu)} = -\frac{1}{au} + \frac{b}{a^2(a+bu)}$$ Step 3: Perform the integration

Step 3: Integrate the simplified integrand

We will now integrate our simplified integrand term by term: $$\int \frac{1}{u^2(a+bu)} du = -\int\frac{1}{au} du + \frac{b}{a^2}\int\frac{1}{a+bu} du$$ Integrating both terms, we get: $$-\frac{1}{a}\ln|u| + \frac{b}{a^2}\ln|a+bu| + C$$ Step 4: Simplify and combine the logarithm terms

Step 4: Simplify the expression by combining the logarithms

Using the logarithmic property of division, we simplify and rewrite the expression as: $$-\frac{1}{a}\ln|u| + \frac{b}{a^2}\ln|a+bu| = -\frac{1}{au} + \frac{b}{a^2}\ln\left|\frac{a+bu}{u}\right| + C$$ Thus, we have derived the formula as requested: $$\int \frac{d u}{u^{2}(a+b u)}=-\frac{1}{au}+\frac{b}{a^{2}} \ln \left|\frac{a+b u}{u}\right|+C$$

Key Concepts

Partial Fraction Decomposition

When faced with a complex rational expression, partial fraction decomposition is a powerful tool that breaks it down into simpler fractions that are easier to integrate. Imagine trying to navigate a bustling city; rather than tackling the entire map at once, we break it down into familiar blocks—partial fractions work similarly for integrating complex expressions. Let's say we have an integrand like \(\frac{1}{u^2(a+bu)}\). We would express it as a sum such as \(\frac{A}{u} + \frac{B}{u^2} + \frac{C}{a+bu}\).

Once set up, the challenge becomes finding the right values for A, B, and C. This is like solving a puzzle where you need to fit the right pieces to complete the picture. Here, equating coefficients and strategically choosing values of 'u' can help us deduce the coefficients. This step is key for the accuracy of the method. Always remember to multiply both sides by the original denominator and solve the resulting equation to determine A, B, and C. Students must practice this step thoroughly to become proficient at identifying the correct coefficients for the decomposition.

Integral Calculus

Integral calculus is one of the twin branches of calculus, partnered with differential calculus, and it's akin to assembling the pieces of a disassembled object. If we consider differentiation as a method to break down, then integration is the method to build up or sum parts to find wholes. In our exercise, once we've used partial fraction decomposition to simplify the integrand, integral calculus comes into play. With simplified fractions like \(\int\frac{1}{au} du\) and \(\int\frac{1}{a+bu} du\), their integration becomes straightforward; think of it as summing smaller, manageable pieces rather than tackling a daunting whole. The goal here is to recognize basic integral forms and apply them successfully. The integration process often involves recognizing patterns that can be directly integrated, for example, integrating a form like \(\int\frac{1}{x} dx\) which gives a logarithmic function.

Logarithmic Integration

Imagine logarithmic integration as learning a new language—a language that allows us to translate certain algebraic expressions into elegant logarithmic forms. In logarithmic integration, we deal with integrals of the form \(\int\frac{1}{x} dx\), which translates to \(\ln|x|+C\). In our exercise, we use logarithmic integration for terms like \(\int\frac{-1}{au} du\) and \(\int\frac{b}{a+bu} du\), resulting in logarithmic expressions with constants. After integrating, we often need to manipulate these logarithms using properties like \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\) to simplify our answer. Students should familiarize themselves with these properties to efficiently handle the logs that occur in such integrals. In this way, the daunting integrals are unraveled into clean, logarithmic statements that can easily be manipulated and interpreted.