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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 8, Problem 8.5.50

Evaluate the integrals in Exercises 39–54.
∫ 1 / (x⁶(x⁵ + 4)) dx

Verified step by step guidance
1
Start by rewriting the integral to clearly see the expression: \(\int \frac{1}{x^{6}(x^{5} + 4)} \, dx\).
Consider a substitution to simplify the integral. Notice that \(x^{5}\) appears inside the parentheses, so let \(u = x^{5} + 4\).
Differentiate \(u\) with respect to \(x\) to find \(du\): \(du = 5x^{4} \, dx\), which implies \(dx = \frac{du}{5x^{4}}\).
Rewrite the integral in terms of \(u\) and \(x\): substitute \(x^{5} + 4\) with \(u\) and \(dx\) with \(\frac{du}{5x^{4}}\). Also, express the remaining powers of \(x\) in terms of \(u\) if possible.
Simplify the integral after substitution and look for further algebraic manipulation or partial fraction decomposition to integrate with respect to \(u\).

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Key Concepts

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

Integration of Rational Functions

This involves integrating functions expressed as the ratio of two polynomials. Techniques often include algebraic manipulation, partial fraction decomposition, or substitution to simplify the integral into manageable parts.
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Intro to Rational Functions

Partial Fraction Decomposition

A method used to break down complex rational expressions into simpler fractions that are easier to integrate. It requires factoring the denominator and expressing the integrand as a sum of simpler rational terms.
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Partial Fraction Decomposition: Distinct Linear Factors

Substitution Method

A technique where a part of the integrand is replaced with a new variable to simplify the integral. It is especially useful when the integral contains a composite function or when the derivative of a function appears within the integrand.
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Euler's Method