Problem 42 - Calculus: One and Several Variables

Question

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

Step-by-Step Solution

Answer

The formula $$\int \frac{d u}{u(a+b u)^{2}}$$ can be derived through the following steps: 1. Perform the substitution \( v = a + bu \) and change du to dv terms. 2. Integrate the resulting expression by applying partial fraction decomposition. 3. Calculate the coefficients A, B, and C. 4. Integrate the decomposed fractions. 5. Convert the expression back to u terms. The final result is: $$\int \frac{du}{u(a+bu)^{2}} = \frac{1}{a(a+bu)} + \frac{1}{a^{2}} \ln \left|\frac{a+bu}{u}\right| + C$$

Step 1: Identify the substitution

First, we identify the function u and its derivative. Here, u is already given, and we are integrating with respect to its derivative, du. We also have the function u(a + bu)^2 in the denominator.

Step 2: Perform the substitution

We substitute u in the integration and perform the integration by substituting du as well. Let's define a new variable v = a + bu. Then the derivative of v with respect to u is given by dv/du = b, or \( dv=b \cdot du \) The integral then becomes: $$ \int \frac{d u}{u(a+b u)^{2}} = \int \frac{1}{u \cdot v^{2}} \cdot \frac{dv}{b} $$

Step 3: Separate constants from the integral

Now, we separate the constants from the integral. In this case, the constant part is 1/b. We get: $$ \frac{1}{b} \int \frac{d v}{u \cdot v^{2}} $$

Step 4: Convert back to u terms

Since we made the substitution, we must now convert back to u terms. From v = a + bu, we can write u in terms of v: $$ u= \frac{v-a}{b} $$ Plugging this back into the integral, we get: $$ \frac{1}{b} \int \frac{d v}{(\frac{v-a}{b}) \cdot v^{2}} $$ Simplifying, this becomes: $$ \int \frac{d v}{(v-a) \cdot v^{2}} $$

Step 5: Apply partial fraction decomposition

To integrate this fraction, we will use partial fraction decomposition. Express the fraction as the sum of two simpler fractions: $$ \frac{1}{(v-a) \cdot v^{2}} = \frac{A}{v-a} + \frac{B}{v} + \frac{C}{v^{2}} $$ Multiplying both sides by \((v-a) \cdot v^{2}\), we get: $$ 1 = A \cdot v^{2} + B \cdot (v-a) \cdot v + C (v-a) $$

Step 6: Find the coefficients A, B, and C

Set v = a to find the coefficient A: $$ 1 = A \cdot a^{2} $$ So, \( A = \frac{1}{a^{2}} \). Now differentiate both sides with respect to v and set v = 0 to find the coefficient B: $$ 0 = 2Aa + B(a-0) - C $$ Solving for B, we get: $$ B = \frac{-C}{a} $$ Finally, set v = 1 to find the coefficient C. Plugging A, and B into the equation, we get: $$ 1 = \frac{1}{a^{2}} + \left(\frac{-C}{a}\right) + C(1-a) $$ Solving for C, we get $$ C = \frac{1}{a} $$

Step 7: Integrate the decomposed fractions

Plugging the coefficients back into the partial fraction decomposition, we have: $$ \int \frac{d v}{(v-a) \cdot v^{2}} = \int \left(\frac{1}{a^{2}} \cdot \frac{d v}{v-a} - \frac{1}{a} \cdot \frac{d v}{v} + \frac{1}{a} \cdot \frac{d v}{v^{2}}\right) $$ Now, integrate each term separately: $$ = \frac{1}{a^{2}} \int \frac{d v}{v-a} - \frac{1}{a} \int \frac{d v}{v} + \frac{1}{a} \int \frac{d v}{v^{2}} $$ The integral of 1/(v-a) is ln|v-a|, the integral of 1/v is ln|v| and the integral of 1/(v^2) is -1/v: $$ = \frac{1}{a^{2}} \ln \left|\frac{v-a}{v}\right| - \frac{1}{a} \cdot \frac{1}{v} + C $$

Step 8: Convert back to u terms

Change back from v to u terms using v = a + bu: $$ \int \frac{d u}{u(a+b u)^{2}} = \frac{1}{a^{2}} \ln \left|\frac{(a+b u)-a}{a+b u}\right| - \frac{1}{a} \cdot \frac{1}{a+b u} + C $$ Finally, simplify the expression: $$ \int \frac{ d u}{u(a+b u)^{2}} = \frac{1}{a(a+b u)} + \frac{1}{a^{2}} \ln \left|\frac{a+b u}{u}\right| + C $$

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in calculus to break down complex rational expressions into simpler ones that are easier to integrate. Since integrating complex fractions directly can be challenging, this method transforms them into a sum of simpler fractions whose antiderivatives are readily known.

The steps involved typically include:

• Factoring the denominator of the rational expression.
• Writing the expression as a sum of fractions with unknown coefficients.
• Determining these coefficients by solving a system of equations that comes from equating the numerators of both sides.

In our exercise, the rational expression \( \frac{1}{(v-a) \cdot v^{2}} \) was decomposed into simpler fractions. This method is particularly useful when dealing with definite integrals, as it simplifies the process of finding the antiderivative.

Definite Integrals in Calculus

Definite integrals play a crucial role in calculus, representing the area under the curve of a function between two points. Unlike indefinite integrals, which give us a general antiderivative plus a constant of integration, definite integrals result in a specific numerical value.

The process of calculating a definite integral involves finding the antiderivative of a function and then evaluating it at the upper and lower limits of integration.

• Antiderivatives provide us with a function F(x) from the derivative f(x).
• Applying the limits to F(x) as per the Fundamental Theorem of Calculus, we get F(b) - F(a), where [a, b] is the interval of integration.

This is fundamentally applying the antiderivative concept to find the total accumulation of a quantity, like area, over an interval defined by the limits of integration.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the inverse operation to differentiation. For a continuous function f(x), its antiderivative is a function F(x) such that F'(x) = f(x). Since differentiation wipes out constant information, the antiderivative includes an arbitrary constant, C.

To find antiderivatives, we rely on a repertoire of integration techniques, including substitution and partial fraction decomposition.

• The substitution method involves changing the variable of integration to simplify the integral.
• Partial fraction decomposition is used when the integrand is a rational function that can be broken into simpler fractions.

Integrating by substitution was applied in our problem, where substitution made it possible to turn a difficult integral into an easier one that could be approached with partial fraction decomposition. Together, these techniques yield a set of simpler antiderivatives that are combined to find the overall integral of the original complex function.