Problem 67 - Calculus Single Variable

Question

Integrate by parts. This will result in an integrand of the form \(P(x) / Q(x)\) where \(P(x)\) and \(Q(x)\) are polynomials with the degree of \(P(x)\) greater than or equal to the degree of \(Q(x)\). Such an integrand is handled by performing polynomial division to put \(P(x) / Q(x)\) into the form \(r(x)+s(x) / Q(x)\) where \(r(x)\) and \(s(x)\) are polynomials with the degree of \(s(x)\) less than the degree of \(Q(x)\) $$ \int_{0}^{1} \ln \left(1+x^{2}\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln(2) - 1 + \frac{\pi}{2} \).

Step 1: Identify Parts for Integration by Parts

The integral \( \int_{0}^{1} \ln(1+x^2) \, dx \) can be considered using integration by parts. Let \( u = \ln(1+x^2) \) and \( dv = dx \). We need to find \( du \) and \( v \).

Step 2: Determine \( du \) and \( v \)

Differentiate \( u \) to get \( du = \frac{2x}{1+x^2} \, dx \). Integrate \( dv = dx \) to get \( v = x \).

Step 3: Apply Integration by Parts Formula

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). Use \( u = \ln(1+x^2) \) and \( v = x \) to get: \[ \int \ln(1+x^2) \, dx = x \ln(1+x^2) - \int x \left( \frac{2x}{1+x^2} \right) \, dx. \]

Step 4: Simplify \( \int x \left( \frac{2x}{1+x^2} \right) \, dx \)

Simplify the integral: \( \int x \left( \frac{2x}{1+x^2} \right) \, dx = \int \frac{2x^2}{1+x^2} \, dx \). This can be rewritten as \( \int (2 - \frac{2}{1+x^2}) \, dx \) by polynomial division.

Step 5: Integrate \( 2 - \frac{2}{1+x^2} \)

Split the integral: \( \int 2 \, dx - \int \frac{2}{1+x^2} \, dx \). Use separate antiderivatives: \( \int 2 \, dx = 2x \) and \( \int \frac{2}{1+x^2} \, dx = 2 \tan^{-1}(x) \).

Step 6: Combine and Evaluate Definite Integral

Combine the results: \( x \ln(1+x^2) - 2x + 2 \tan^{-1}(x) \). Evaluate from 0 to 1: \[[ x \ln(1+x^2) - 2x + 2 \tan^{-1}(x) ]_0^1 \].

Step 7: Final Calculation

Calculate: \( [1 \ln(2) - 2 + 2 \frac{\pi}{4}] - [0 - 0 + 0] = \ln(2) - 1 + \frac{\pi}{2} \).

Key Concepts

Polynomial Division

When dealing with integrals like \( \int \frac{P(x)}{Q(x)} \, dx \), where both \(P(x)\) and \(Q(x)\) are polynomials, polynomial division is a critical step. Polynomial division helps simplify the integrand into a form that is easier to integrate.

In the context of the given exercise, the polynomial division is used after applying the integration by parts technique. This step converts the complex fraction \( \frac{2x^2}{1+x^2} \) into a simpler form, such as \(2 - \frac{2}{1+x^2}\).

The goal is to express the division of the polynomials as a polynomial plus a simpler rational function. This transformation is essential because it often leads to components that are much more straightforward to integrate, such as constants or other identifiable antiderivatives.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions whose derivative is the original function given in the integrand. An understanding of antiderivatives is crucial to solving integrals, including in integration by parts.

In step 5 of the original exercise, we need the antiderivatives of two simpler functions: \(\int 2 \, dx\) and \(\int \frac{2}{1+x^2} \, dx\).

• For \(\int 2 \, dx\), the antiderivative is simply \(2x\), as the derivative of \(2x\) is 2.
• For \(\int \frac{2}{1+x^2} \, dx\), the antiderivative is \(2 \tan^{-1}(x)\), because the derivative of \(\tan^{-1}(x)\) is \(\frac{1}{1+x^2}\).

Recognizing these standard antiderivatives is key to efficiently working through the process and applying them to evaluate definite integrals.

Definite Integrals

A definite integral is an integral expressed with upper and lower limits of integration. This type of integral represents the net area under a curve within a given interval. The definite integral from a to b of function f(x), written as \(\int_a^b f(x) \, dx\), is solved by finding an antiderivative and then evaluating it at the upper limit minus the lower limit.

In our example, the integration by parts and polynomial division techniques help us arrive at the expression \( x \ln(1+x^2) - 2x + 2 \tan^{-1}(x) \).

To find the definite integral \( \int_{0}^{1} \ln(1+x^2) \, dx \), we evaluate the antiderivative at the upper limit (x = 1) and subtract its value at the lower limit (x = 0).

This final calculation results in an answer of \( \ln(2) - 1 + \frac{\pi}{2} \), showcasing the power of integrating with methods like integration by parts and simplifying using techniques such as polynomial division. This not only highlights the area interpretation but also builds a deeper understanding of the process involved in reaching the result.