## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 38

#### Answer

$$3\ln \frac{4}{3}$$

#### Work Step by Step

Given $$\int_{8}^{27} \frac{d x}{x+x^{2 / 3}}$$ Let $$u^3=x\ \ \ \ \ \ 3 u^2du =dx$$ At $x=8\ \ \to \ u=2$ and at $x=27\ \ \to \ u=3$, so we have \begin{align*} \int_{8}^{27} \frac{d x}{x+x^{2 / 3}}&=\int_{2}^{3} \frac{3 u^2du}{u^3+u^{2 }}\\ &= \int_{2}^{ 3} \frac{3 du}{u +1}\\ &= 3\ln|u+1|\bigg|_{2}^{ 3}\\ &= 3\ln| 3+1|-3\ln |2+1|\\ &=3\ln \frac{4}{3} \end{align*}

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