Problem 298 - Calculus Volume 1

Question

For the following exercises, find the indefinite integral. $$\int \frac{d t}{3 t}$$

Step-by-Step Solution

Answer

\( \int \frac{d t}{3 t} = \frac{1}{3} \ln |t| + C.\)

Step 1: Recognize the Integrand Format

The integrand can be recognized as a fraction where the numerator is simply the differential \(dt\) and the denominator is \(3t\). This suggests the form \(\frac{1}{3} \int \frac{1}{t} \, dt\).

Step 2: Pull Out Constant Factor

Observe that \(\frac{1}{3}\) is a constant factor. We can bring this constant outside the integral as follows: \(\frac{1}{3} \int \frac{1}{t} \, dt\).

Step 3: Integrate Using Basic Logarithm Rule

Recall the integral rule for \(\int \frac{1}{x} \, dx = \ln |x| + C\). Apply this rule to integrate:\[\frac{1}{3} \int \frac{1}{t} \, dt = \frac{1}{3} \ln |t| + C,\]where \(C\) is the constant of integration.

Step 4: Final Step: Present the Solution

Combine the results to present the final solution:\[\int \frac{d t}{3 t} = \frac{1}{3} \ln |t| + C.\]

Key Concepts

Integration by Substitution

Integration by substitution is a method used when the integrand is a composite function. This technique typically simplifies the integral by making a substitution to transform the integrand into an easier form. To apply substitution:

• Identify a part of the integrand as a new variable, usually denoted as "\(u\)".
• Substitute this new variable into the integrand, replacing the original variable.
• Once the integral is in terms of \(u\), integrate with respect to \(u\).
• After integration, substitute back the original variable to express the final result.

This is particularly helpful for integrals that involve products or compositions of functions, where one function is the derivative of another. However, in the original problem \(\int \frac{d t}{3 t}\), integration by substitution isn't necessary, as the integrand is already in a straightforward form for logarithmic integration.

Logarithmic Integration

Logarithmic integration is applicable when encountering an integrand of the form \(\frac{1}{x}\). This is because there is a well-known rule for integrating this form:

• The integral of \(\frac{1}{x}\) is \(\ln |x| + C\), where \(C\) is the constant of integration.

In the original exercise, the integrand is \(\frac{1}{t}\) multiplied by a constant factor, \(\frac{1}{3}\). This can be rewritten as \(\frac{1}{3} \int \frac{1}{t} \, dt\). The integration rule is straightforward, allowing us to apply the logarithmic rule directly:

• Integrate \(\frac{1}{t}\) to get \(\ln |t|\),
• Then multiply by the constant \(\frac{1}{3}\).

Therefore, the integral of \(\frac{d t}{3 t}\) is \(\frac{1}{3} \ln |t| + C\).

Constant Factor in Integration

A constant factor in an integral can be placed outside the integral sign, making the process simpler. This technique relies on a property of definite and indefinite integrals, allowing algebraic constants to be factored out:

• If you have an integral of the form \(\int k \cdot f(x) \, dx\), it can be rewritten as \(k \cdot \int f(x) \, dx\), where \(k\) is a constant.

In the exercise, the given integrand is \(\frac{1}{3} \cdot \frac{1}{t}\). The constant factor \(\frac{1}{3}\) was taken out of the integral, which simplifies the integral into \(\frac{1}{3} \cdot \int \frac{1}{t} \, dt\). This factorization simplifies the integration process, and only the core function \(\frac{1}{t}\) needs to be integrated, while the constant remains untouched outside multiplying the resultant integral.