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Calculus: Early Transcendentals

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

x +1
「マ-x-6
dx.
х2 — х —

Expert Solution

Step 1

The integral is $\int \frac{x + 1}{x^{2} - x - 6} d x$ .

Factorize the denominator.

$\begin{array}{rcl} x^{2} - x - 6 & = & x^{2} - 3 x + 2 x - 6 \\ = & x (x - 3) + 2 (x - 3) \\ = & (x - 3) (x + 2) \end{array}$

Thus, the integral becomes.

$\int \frac{x + 1}{(x - 3) (x + 2)} d x$

Step 2

Now consider the function $\frac{x + 1}{(x - 3) (x + 2)}$ .

Split the terms.

$\begin{array}{rcl} \frac{x + 1}{(x - 3) (x + 2)} & = & \frac{A}{x - 3} + \frac{B}{x + 2} \\ = & \frac{A (x + 2) + B (x - 3)}{(x - 3) (x + 2)} \\ = & \frac{A x + 2 A + B x - 3 B}{(x - 3) (x + 2)} \\ = & \frac{x (A + B) + (2 A - 3 B)}{(x - 3) (x + 2)} \end{array}$

On comparing the coefficients of both sides of the above equation,

$\begin{array}{rcl} A + B & = & 1 . . . . . . (1) \\ 2 A - 3 B & = & 1 . . . . . . (2) \end{array}$

Multiply to equation (1) and then add it with equation (2).

$\begin{array}{rcl} 5 A & = & 4 \\ A & = & \frac{4}{5} \end{array}$

Substitute $\frac{4}{5}$ for A equation (1).

$\begin{array}{rcl} \frac{4}{5} + B & = & 1 \\ B & = & 1 - \frac{4}{5} \\ = & \frac{1}{5} \end{array}$

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