Calculus (3rd Edition)

Published by W. H. Freeman

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 34

Answer

$$\frac{1}{7}\ln \frac{24}{10}$$

Work Step by Step

Given $$\int_{4}^{6} \frac{d t}{(t-3)(t+4)}$$ Since \begin{aligned} \frac{1}{(t-3)(t+4)} &=\frac{A}{(t-3)}+\frac{B}{(t+4)} \\ &=\frac{A(t+4)+B(t-3)}{(t-3)(t+4)} \\ 1 &=A(t+4)+B(t-3) \end{aligned} \begin{align*} \text{at }\ t&= -4\ \ \ \ \ \ \ A=\frac{1}{7}\\ \text{at }\ t&= 3\ \ \ \ \ \ \ B=\frac{-1}{7} \end{align*} Then \begin{aligned} \int_{4}^{6} \frac{d t}{(t-3)(t+4)} &=\frac{1}{7} \int_{4}^{6} \frac{1}{(t-3)} d t-\frac{1}{7} \int_{4}^{6} \frac{1}{(t+4)} d t \\ &=\frac{1}{7} \ln |t-3|-\frac{1}{7} \ln |t+4|\bigg|_{4}^{6}\\ &= \frac{1}{7} \ln |3|-\frac{1}{7} \ln |10|+\frac{1}{7} \ln |8|\\ &=\frac{1}{7}\ln \frac{24}{10} \end{aligned}

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