Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int \arctan x d x $$

Short answer

The integral \( \int \arctan x \, dx = x \cdot \arctan x - \frac{1}{2} \ln |1 + x^2| + C \).

Step-by-step solution

Choose Functions for Integration by Parts

In integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). For the integral \( \int \arctan x \, dx \), choose \( u = \arctan x \) because its derivative is simpler, and \( dv = dx \), so that \( v = x \).

Differentiate and Integrate

Differentiate \( u \): \( du = \frac{1}{1+x^2} \, dx \). Integrate \( dv \): \( v = x \). Now we have the components needed for substitution into the integration by parts formula.

Apply Integration by Parts Formula

Substitute the chosen \( u \), \( du \), \( v \), and \( dv \) into the formula: \[ \int \arctan x \, dx = x \cdot \arctan x - \int x \cdot \frac{1}{1+x^2} \, dx \].

Simplify the Remaining Integral

Focus on the remaining integral \( \int \frac{x}{1+x^2} \, dx \). Notice that the denominator is the derivative of the numerator plus a constant, suggesting the use of substitution.

Substitution for Remaining Integral

Let \( w = 1 + x^2 \), then \( dw = 2x \, dx \), so \( x \, dx = \frac{1}{2} dw \). This transforms the integral to \( \frac{1}{2} \int \frac{1}{w} \, dw \).

Solve the Substituted Integral

The integral \( \frac{1}{2} \int \frac{1}{w} \, dw \) becomes \( \frac{1}{2} \ln |w| + C \). Substitute back \( w = 1 + x^2 \) to get \( \frac{1}{2} \ln |1 + x^2| \) as the antiderivative.

Combine Results

Substitute back into Step 3's result: \( x \cdot \arctan x - \frac{1}{2} \ln |1 + x^2| + C \). This is your solution after applying integration by parts.

Key concepts

calculus

Calculus is a field of mathematics that focuses on rates of change and the accumulation of quantities. It comprises two primary branches: **differential calculus** and **integral calculus**. Differential calculus is concerned with the concept of a derivative, essentially describing how something changes. On the other hand, integral calculus is focused on the concept of integration, which is about finding the quantity where the derivative (or rate of change) is known.
A fundamental concept in integral calculus is being able to find the area under curves or, more technically, finding antiderivatives. One influential technique for solving integrals, especially those involving products of functions, is **integration by parts**. This method is a direct consequence of the product rule for differentiation and is especially useful for integrating products of an easy-to-differentiate function and one easy-to-integrate function.
The formula used in integration by parts is given by:

• \[ \int u \, dv = uv - \int v \, du \]

Using this, we break down an integral into parts that are easier to handle individually, as seen in the original exercise where \( u = \arctan x \) and \( dv = dx \). Calculus provides the toolset for solving problems where finding these antiderivatives are necessary.

trigonometric integrals

Trigonometric integrals are integrals that involve trigonometric functions like sine, cosine, tangent, etc. In the exercise involving \( \int \arctan x \, dx \), although \( \arctan x \) is not directly one of the basic trigonometric functions, it is related through its derivative and application in various problems that involve calculus.
When you're handling integrals with trigonometric functions, it often involves using identities to simplify the process. For example:

• Pythagorean identities, like \( \sin^2 x + \cos^2 x = 1 \)
• Sum and difference identities
• Double angle formulas

These identities can turn complex integrals into something more manageable.\
In the provided solution, the trigonometric nature of \( \arctan x \), although not needing identities directly, plays a crucial role in defining the derivatives and understanding its behavior during integration by parts. Recognizing when and how to apply these identities is key in making integrals simpler and solvable.

substitution method

The substitution method, also known as u-substitution, is a powerful technique in calculus for finding integrals. It's a way to simplify an integral by changing variables to make the integration process easier. Essentially, you choose a new variable \( u \) that simplifies the integral, replacing a part of the integrand and its differential.
In our exercise, substitution was applied to simplify the integral \( \int \frac{x}{1+x^2} \, dx \). By choosing \( w = 1 + x^2 \), the substitution turns the integral into something far simpler. The derivative of this new variable, \( dw = 2x \, dx \), allowed us to write \( x \, dx \) as \( \frac{1}{2} dw \), leading to an easier integral \( \frac{1}{2} \int \frac{1}{w} \, dw \), which is a straightforward logarithmic integral.

• The substitution method is particularly useful when the integral includes a chain of functions.
• It often simplifies integrals involving products, compositions, or other complex expressions.

Substitution allows us to transform complex integrals into forms that are immediately solvable by recognized methods, such as the natural logarithm in our solution.