## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 39

#### Answer

$$\frac{2}{\sqrt{a}}\tan^{-1}\frac{\sqrt{x}}{\sqrt{a}}+C,\ \ \ \ \ \ a\neq0$$ $$\frac{-2}{\sqrt{x}}+C,\ \ \ \ \ \ a=0$$

#### Work Step by Step

Given $$\int \frac{d x}{x^{3 / 2}+a x^{1 / 2}}$$ Let $$x=u^2 \ \ \ \ \ \ dx=2udu$$ Then for $a\neq 0$ \begin{align*} \int \frac{d x}{x^{3 / 2}+a x^{1 / 2}}&=\int \frac{2ud u}{u^{3 }+a u}\\ &= \int \frac{2 d u}{u^{2 }+a }\\ &= \frac{2}{\sqrt{a}}\tan^{-1}\frac{u}{\sqrt{a}}+C\\ &= \frac{2}{\sqrt{a}}\tan^{-1}\frac{\sqrt{x}}{\sqrt{a}}+C \end{align*} For $a=0$ \begin{align*} \int \frac{d x}{x^{3 / 2}}&=\frac{-2}{\sqrt{x}}+C \end{align*}

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