Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 35


$$-\frac{1}{49} \ln |t-3|-\frac{1}{7(t-3)}+\frac{1}{49} \ln |t+4|+C$$

Work Step by Step

Given $$ \int \frac{d t}{(t-3)^{2}(t+4)}$$ Since \begin{aligned} \frac{1}{(t-3)^{2}(t+4)} &=\frac{A}{(t-3)}+\frac{B}{(t-3)^{2}}+\frac{C}{(t+4)} \\ &=\frac{A(t-3)(t+4)+B(t+4)+C(t-3)^{2}}{(t-3)^{2}(t+4)} \\ 1 &=A(t-3)(t+4)+B(t+4)+C(t-3)^{2} \end{aligned} Then \begin{align*} \text{at } x&=3 \ \ \ \ \to A=\frac{1}{7} \\ \text{at } x&= -4\ \ \ \ \to C= \frac{1}{49}\\ \text{at } x&= 04\ \ \ \ \to B= \frac{1}{49} \end{align*} Then \begin{aligned} \int \frac{d t}{(t-3)^2(t+4)} &=-\frac{1}{49} \int \frac{1}{(t-3)} d t+\frac{1}{7} \int \frac{1}{(t-3)^{2}} d t+\frac{1}{49} \int \frac{1}{(t+4)} d t \\ &=-\frac{1}{49} \int \frac{1}{(t-3)} d t+\frac{1}{7} \int(t-3)^{-2} d t+\frac{1}{49} \int \frac{1}{(t+4)} d t \\ &=-\frac{1}{49} \ln |t-3|+\frac{1}{7} \cdot \frac{(t-3)^{-1}}{(-1)}+\frac{1}{49} \ln |t+4|+C \\ &=-\frac{1}{49} \ln |t-3|-\frac{1}{7(t-3)}+\frac{1}{49} \ln |t+4|+C \end{aligned}
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