Problem 68 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \arctan x d x=\) (A) \(x \arctan x-\ln \left(1+x^{2}\right)+C\) (B) \(x \arctan x+\ln \left(1+x^{2}\right)+C\) (C) \(x \arctan x+\frac{1}{2} \ln \left(1+x^{2}\right)+C\) (D) \(x \arctan x-\frac{1}{2} \ln \left(1+x^{2}\right)+C\)

Step-by-Step Solution

Answer

The correct answer is (D) \( x \arctan x - \frac{1}{2} \ln(1+x^2) + C \).

Step 1: Identify the Integration Technique

The integral \( \int \arctan x \, dx \) requires integration by parts. We choose \( u = \arctan x \) and \( dv = dx \). This technique is useful when an integral involves a product of functions or can be simplified using the products.

Step 2: Differentiate and Integrate Parts

Apply the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).- Differentiate \( u = \arctan x \): \( du = \frac{1}{1+x^2} \, dx \).- Integrate \( dv = dx \): \( v = x \).

Step 3: Substitute and Simplify

Substitute into the integration by parts formula: \[ \int \arctan x \, dx = x \arctan x - \int x \cdot \frac{1}{1+x^2} \, dx \].

Step 4: Solve the Remaining Integral

The remaining integral is \( \int \frac{x}{1+x^2} \, dx \). Recognize this as a natural logarithm form, where \( \frac{d}{dx}[\ln(1+x^2)] = \frac{2x}{1+x^2} \). Thus, simplify to get: \[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln(1+x^2) + C \].

Step 5: Combine Results

Substitute back the result into the equation obtained from integration by parts: \[ x \arctan x - \frac{1}{2} \ln(1+x^2) + C \].

Step 6: Compare with Options and Choose the Correct One

Compare with the given options. The result matches option (D): \[ x \arctan x - \frac{1}{2} \ln(1+x^2) + C \].

Key Concepts

Understanding Arctan Integration

Arctan, or the inverse tangent function, can be tricky to integrate directly. The integral \( \int \arctan x \; dx \) poses a challenge because it combines an inverse trigonometric function, which doesn’t have a straightforward antiderivative. This is where integration by parts becomes valuable.

• In our exercise, using integration by parts means we let \( u = \arctan x \) and \( dv = dx \).
• The strategy behind this choice relies on simplifying the integral into manageable parts by differentiating one part and integrating the other.

This technique allows us to transform the integral into a more solvable form, as seen in the step-by-step solution. By leveraging the integration by parts formula, the complexity of dealing with \( \arctan x \) directly is reduced. Good knowledge of arctan behaviors and properties helps in simplifying such integrals.

Exploring Logarithmic Integration

Logarithmic integration often arises when integrating expressions of the form \( \int \frac{1}{x} \, dx \), forming the natural logarithm, \( \ln(x) \). In context, integrating the expression \( \int \frac{x}{1+x^2} \, dx \) taps into this concept.

• Notice that \( \frac{x}{1+x^2} \) looks similar to \( \frac{du}{u} \), a derivative relationship involving the natural logarithm.
• Here, considering \( 1 + x^2 \) as \( u \) effectively transforms the expression into a derivative form, \( \frac{d}{dx}[\ln(1+x^2)] \).

To solve \( \int \frac{x}{1+x^2} \, dx \), it is key to recognize this as half the derivative of \( \ln(1+x^2) \). Thus, its integral is \( \frac{1}{2}\ln(1+x^2) + C \). This elegant solution underscores the power of understanding derivative-integration relationships with logarithmic functions.

AP Calculus Integration Problems

Solving integration problems in AP Calculus involves recognizing patterns and applying appropriate techniques. The integral explored here uses integration by parts, which is a cornerstone method in calculus for solving products of functions. Through these problems, students are expected to:

• Identify when specific rules or methods, like integration by parts or substitution, are suitable.
• Understand inverse trigonometric integrals and logarithmic relationships.

Mastering these concepts requires practice. Knowing how to break down and solve an integral can be rewarding, especially when it involves more complex functions like \( \arctan x \).Additionally, AP Calculus students benefit from accumulating a toolkit of techniques. This includes knowing when to use integration by parts to simplify integrals that cannot be solved immediately. Such problems might first appear daunting, but with practice, they reveal underlying structures that make them manageable.