Problem 10 - Calculus: Early Transcendental Functions

Question

Find the general antiderivative. $$\int \frac{x+2 x^{3 / 4}}{x^{5 / 4}} d x$$

Step-by-Step Solution

Answer

Therefore, the general antiderivative of the given function is \(\frac{4}{3} x^{4/3} - 2 ln|x| + c\)

Step 1: Simplify the integrand

The integrand can be simplified by dividing each term of the numerator by the denominator: \(\int \frac{x+2 x^{3 / 4}}{x^{5 / 4}} d x = \int \frac{x}{x^{5 / 4}} dx + \int \frac{2 x^{3 / 4}}{x^{5 / 4}} dx = \int x^{5/4 - 1} dx + \int 2 x^{3/4 - 5/4} dx = \int x^{-1 / 4} dx + \int 2x^{-1} dx\).

Step 2: Integrate terms separately

Integrate each term separately using the power rule for integration which states that the integral of \(x^n dx = {1/(n+1)} x^{n+1}\): \[\int x^{-1 / 4} d x = \frac{4}{3} x^{4/3} + c_1\] \[\int 2x^{-1} dx = 2 \int x^{-1} dx = -2 ln|x| + c_2\] where \(c_1\) and \(c_2\) are constants of integration.

Step 3: Combine the results

Combine the separate integral results and their respective constant of integration into a single expression: \(\frac{4}{3} x^{4/3} - 2 ln|x| + c\), where \(c = c_1 + c_2\), which we can't determine based on the information given, so it is generally written as + c.

Key Concepts

Integrand Simplification

Integrand simplification is the process of breaking down a complex expression into simpler parts for easier integration. In the given exercise, we start with the integrand \( \frac{x+2x^{3/4}}{x^{5/4}} \). Instead of trying to integrate this as it is, we simplify each term separately by dividing the terms in the numerator by the denominator.

Let's see how this works step-by-step:

• The term \( \frac{x}{x^{5/4}} \) simplifies to \( x^{1-5/4} = x^{-1/4} \).
• The term \( \frac{2x^{3/4}}{x^{5/4}} \) simplifies to \( 2x^{3/4-5/4} = 2x^{-1} \).

Now, the original complex integrand is much simpler: \( x^{-1/4} + 2x^{-1} \). This simplification allows us to apply standard integration techniques more easily.

Breaking down the integrand in this step ensures that you approach the integration process more effectively, paving the way for smoother calculation and minimized room for error.

Power Rule for Integration

The power rule is a straightforward yet essential tool in calculus for finding antiderivatives of expressions in the form \( x^n \). This rule states that when integrating \( x^n \), you add 1 to the exponent and divide by the new exponent. The general formula is:

• \( \int x^n \, dx = \frac{1}{n+1} \, x^{n+1} + C \)
• Note: The formula is applicable only when \( n eq -1 \).

In our exercise, we apply the power rule separately to the simplified integrand terms:

• For \( \int x^{-1/4} \, dx \), the power rule gives us \( \frac{1}{-1/4 + 1} \, x^{3/4} = \frac{4}{3} \, x^{4/3} \).
• \( 2x^{-1} \) requires integration using the known result of \( \int x^{-1} \, dx = \ln|x| \), resulting in \( -2 \ln|x| \).

These calculations show how directly applying the power rule guides us toward finding antiderivatives.

Having a strong understanding of the power rule helps in quick simplification, avoiding potential pitfalls as you work through integrations involving polynomial-like expressions.

Constants of Integration

In indefinite integration, constants of integration play a crucial role. When finding the antiderivative of a function, you are essentially solving for a family of functions. The constant of integration \( C \) represents any constant value that could be added to the antiderivative. This is because the derivative of a constant is zero, leaving the original function unchanged when derived.

In our solution:

• The integration of \( x^{-1/4} \) results in \( \frac{4}{3} x^{4/3} + c_1 \).
• The integration of \( 2x^{-1} \) results in \( -2 \ln|x| + c_2 \).

When combining these, the constants \( c_1 \) and \( c_2 \) simply merge into a single constant, \( c \), since they are arbitrary constants. In practice, we often just write \( + C \) at the end of the final antiderivative.

Recognizing the importance of constants allows for an accurate expression of all possible solutions when performing indefinite integration.