The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀^∞ (16 tan⁻¹x dx) / (1 + x²)

Verified step by step guidance

1. Recognize that the integral is an improper integral with the upper limit approaching infinity, so we will consider the limit as the upper bound tends to infinity: \(\int_0^{\infty} \frac{16 \tan^{-1}(x)}{1 + x^2} \, dx = \lim_{t \to \infty} \int_0^t \frac{16 \tan^{-1}(x)}{1 + x^2} \, dx\).
2. Identify a suitable substitution or integration technique. Notice that the derivative of \(\tan^{-1}(x)\) is \(\frac{1}{1 + x^2}\), which appears in the denominator. This suggests using integration by parts with \(u = \tan^{-1}(x)\) and \(dv = \frac{16}{1 + x^2} dx\).
3. Set up integration by parts: let \(u = \tan^{-1}(x)\) so that \(du = \frac{1}{1 + x^2} dx\), and let \(dv = \frac{16}{1 + x^2} dx\) so that \(v = 16 \tan^{-1}(x)\) (since the integral of \(\frac{1}{1 + x^2}\) is \(\tan^{-1}(x)\)).
4. Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Substitute the expressions for \(u\), \(v\), \(du\), and \(dv\) to rewrite the integral.
5. Simplify the resulting expression and evaluate the limit as \(t \to \infty\). Use the known limits of \(\tan^{-1}(x)\) as \(x\) approaches infinity and zero to find the value of the integral.

Key Concepts

Improper Integrals

Improper integrals involve integration over an infinite interval or integrands with infinite discontinuities. To evaluate them, one typically takes limits to handle the infinite bounds or singularities, ensuring the integral converges to a finite value.

Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. It transforms the integral of a product of functions into simpler integrals, often making complex integrals more manageable, especially when one function simplifies upon differentiation.

Inverse Trigonometric Functions and Their Derivatives

Inverse trigonometric functions like arctan(x) have well-known derivatives, e.g., d/dx [arctan(x)] = 1/(1 + x²). Recognizing these derivatives helps in integration, especially when the integrand includes inverse trig functions combined with rational expressions.