Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.5 The Method of Partial Fractions - Exercises - Page 423: 33

Answer

$$2 \ln |x-1|+\frac{1}{2} \ln \left|x^{2}+1\right|-3 \tan ^{-1} x+C$$

Work Step by Step

Given $$\int \frac{\left(3 x^{2}-4 x+5\right) d x}{(x-1)\left(x^{2}+1\right)}$$ Since \begin{align*} \frac{1}{(x-1)\left(x^{2}+1\right)}&=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+1}\\ &=\frac{A(x^{2}+1)+B(x-1)}{(x-1)\left(x^{2}+1\right)}\\ 1&=A(x^{2}+1)+B(x-1) \end{align*} \begin{align*} \text{at } x&=1 \ \ \ \ \to A=2 \\ \text{at } x&= 0\ \ \ \ \to C=-3\\ \text{at } x&= -1\ \ \ \ \to B =-1\\ \end{align*} Hence $B=-1,\ \ \ C=0$ and \begin{aligned} \int \frac{3 x^{2}-4 x+5}{(x-1)\left(x^{2}+1\right)} d x &=\int \frac{2}{x-1} d x+\int \frac{x-3}{x^{2}+1} d x \\ &=\int \frac{2}{x-1} d x+\int \frac{\frac{1}{2} \cdot 2 x}{x^{2}+1} d x-\int \frac{3}{x^{2}+1} d x \\ &=2 \ln |x-1|+\frac{1}{2} \ln \left|x^{2}+1\right|-3 \tan ^{-1} x+C \end{aligned}
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