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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.6.111

Verify the integration formulas in Exercises 111–114.
111. ∫ (arctan x) / x² dx = ln x - 1/2 ln(1 + x²) - arctan x / x + C

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1
Identify the integral to verify: \(\int \frac{\arctan x}{x^{2}} \, dx\).
Consider using integration by parts. Let \(u = \arctan x\) and \(dv = \frac{1}{x^{2}} dx\).
Compute the derivatives and integrals needed: \(du = \frac{1}{1 + x^{2}} dx\) and \(v = -\frac{1}{x}\).
Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\), which gives \(-\frac{\arctan x}{x} - \int \left(-\frac{1}{x} \cdot \frac{1}{1 + x^{2}}\right) dx\).
Simplify the remaining integral \(\int \frac{1}{x(1 + x^{2})} dx\) by using partial fraction decomposition or substitution, then combine all parts to match the given formula.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and follows the formula ∫u dv = uv - ∫v du. Choosing appropriate u and dv is crucial to simplify the integral effectively.
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Derivative and Integral of Inverse Trigonometric Functions

Understanding the derivatives and integrals of inverse trigonometric functions like arctan(x) is essential. For example, d/dx [arctan x] = 1/(1 + x²). This knowledge helps in manipulating integrals involving arctan(x) and recognizing patterns during integration.
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Logarithmic Integration and Simplification

Integrals often result in logarithmic expressions, especially when integrating rational functions. Recognizing when to rewrite expressions using logarithm properties, such as ln(a) - ln(b) = ln(a/b), aids in simplifying the final answer and verifying given formulas.
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