Question

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

Accepted Answer

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 	an$$^{-1}$$(x)}{1 + x^2$} \, dx = \lim_{t 	o \infty} \$$int_0$$^t \frac{16 	an$$^{-1}$$(x)}{1 + x^2$} \, dx. $$Identify a suitable substitution or integration technique. Notice that the derivative of $$	an$$^{-1}$$(x) is$$ $$\frac{1}{1 + x^2$}$$, which appears in the denominator. This suggests using integration by parts with $$u = 	an$$^{-1}$$(x)$$ and $$dv = \frac{16}{1 + x^2$} dx. $$Set up integration by parts: let $$u = 	an$$^{-1}$$(x) so $$that $$du = \frac{1}{1 + x^2$} dx$$, and let $$dv = \frac{16}{1 + x^2$} dx so $$that $$v = 16 	an$$^{-1}$$(x) ($$since the integral of $$\frac{1}{1 + x^2$} is$$ $$	an$$^{-1}$$(x)). $$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. Simplify the resulting expression and evaluate the limit as $$t 	o \infty. $$Use the known limits of $$	an$$^{-1}$$(x) as$$ $$x$$ approaches infinity and zero to find the value of the integral.

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

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Key Concepts

Improper Integrals

Integration by Parts

Inverse Trigonometric Functions and Their Derivatives