## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - 8.5 The Method of Partial Fractions - Exercises - Page 423: 16

#### Answer

$$3\left(\ln |x|-\ln |x+1|+\frac{1}{(x+1)}\right)+C$$

#### Work Step by Step

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

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