Problem 7 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \frac{2 d u}{1+3 u}=\) (A) \(\frac{2}{3} \ln |1+3 u|+C\) (B) \(-\frac{1}{3(1+3 u)^{2}}+C\) (C) \(2 \ln |1+3 u|+C\) (D) \(\frac{3}{(1+3 u)^{2}}+C\)

Step-by-Step Solution

Answer

The answer is choice (A).

Step 1: Identify the Integral Type

The integral is of the form \(\int \frac{2 \, du}{1+3u}\), which suggests using a substitution or recognizing it as a natural logarithm form \(\int \frac{1}{x} \, dx = \ln |x| + C\).

Step 2: Make a Substitution

Let \(x = 1 + 3u\), then \(dx = 3 \, du\). We need to express \(du\) in terms of \(dx\), which gives \(du = \frac{1}{3}dx\).

Step 3: Substitute into the Integral

Replace \(1+3u\) by \(x\) and \(du\) by \(\frac{1}{3}dx\) in the integral, thus \(\int \frac{2 \, du}{1+3u} = \int \frac{2}{x} \cdot \frac{1}{3}dx = \frac{2}{3} \int \frac{1}{x}\, dx\).

Step 4: Evaluate the Integral

The integral \(\int \frac{1}{x} \, dx = \ln |x| + C\). Therefore, \(\frac{2}{3} \int \frac{1}{x} \, dx = \frac{2}{3} \ln |x| + C\).

Step 5: Back Substitute and Finalize the Solution

Since \(x = 1 + 3u\), replace \(x\) with \(1 + 3u\) to obtain the original variable: \(\frac{2}{3} \ln |1+3u| + C\). This matches choice (A).

Key Concepts

Integration Techniques

When tackling calculus integrals, choosing the right integration technique is essential. Different forms of expressions can suggest different methods of integration. Here are some common techniques often used:

• Basic Antiderivatives: If the integral directly matches a known antiderivative, you can apply it directly.
• Substitution: Useful for integrals that resemble a function and its derivative, allowing simplification.
• Integration by Parts: Good for integrating products of functions, applying the formula \( \int u \, dv = uv - \int v \, du \).
• Partial Fractions: Breaking complex rational expressions into simpler fractions before integrating.
• Trigonometric Substitutions: Useful for expressions involving square roots of quadratic polynomials.

These techniques can turn a seemingly complicated problem into a manageable one, provided you recognize the form of integrals you're dealing with. It’s like solving a puzzle, where recognizing patterns and familiar forms is key to choosing the right tool.

Substitution Method

The substitution method, a powerful integration technique, is especially useful when dealing with integrals that involve composite functions. Here's how it works:

• Identify the Inner Function: Look for a function inside the integral that, if substituted, will simplify the integral.
• Substitute Variables: Introduce a new variable, say \(x\), that simplifies the inner function, allowing you to rewrite the integral in terms of \(x\).
• Express \(du\) with \(dx\): Calculate \(dx\) in terms of \(du\), ensuring all parts of the integral are replaced.
• Integrate: Perform the integration with respect to the new variable.
• Back Substitute: Replace the substituted variable with the original variable to complete the process.

By making the problem simpler, this method can transform a complex integral into one of a basic form. It's a little like changing a tough language problem into a simple math one that you already know how to solve.

Natural Logarithm Integration

Integrating functions that result in natural logarithms is common in calculus, especially when dealing with rational functions. The integral of the form \(\int \frac{1}{x} \, dx\) results in \(\ln |x| + C\). This principle is crucial when:

• Recognizing Forms: Look for integrals where the numerator is the derivative of the denominator.
• Simplifying using Substitution: Use substitution to transform a more complex integral into this form.

In the original exercise, after substitution, the integral simplified to \(\int \frac{1}{x} \, dx\), resulting in \(\ln |x|\). This straightforward relationship makes the natural logarithm integration a handy tool for unraveling certain integrals. It's akin to using a recipe you know well, where the steps become easy once you recognize the form.