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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.6.20

Use the table of integrals at the back of the text to evaluate the integrals in Exercises 1–26.
∫ tan^(-1)(x) / x² dx

Guida verificata passo dopo passo
1
Recognize that the integral is \( \int \frac{\tan^{-1}(x)}{x^2} \, dx \), where \( \tan^{-1}(x) \) is the inverse tangent function, also written as \( \arctan(x) \).
Consider using integration by parts, since the integrand is a product of functions: one involving \( \arctan(x) \) and the other involving \( \frac{1}{x^2} \). Set \( u = \arctan(x) \) and \( dv = \frac{1}{x^2} dx \).
Compute \( du \) and \( v \): - \( du = \frac{1}{1+x^2} dx \) because the derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \). - \( v = \int x^{-2} dx = -x^{-1} = -\frac{1}{x} \).
Apply the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Substitute the expressions for \( u, v, du \) to rewrite the integral.
Simplify the resulting integral and evaluate it using the table of integrals if necessary, focusing on the integral \( \int \frac{1}{x(1+x^2)} dx \) that appears after substitution.

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