Evaluate the integral1xp-9x² + 6x – 8by first completing the square and using the substitutionu = 3x + 1.

Answered: Evaluate the integral 1 xp- 9x² + 6x –…

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

Evaluate the integral
1
xp-
9x² + 6x – 8
by first completing the square and using the substitution
u = 3x + 1.

Expert Solution

Step 1

The given integral is:

$\int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}} d x$

To evaluate this integral, first, make the square by adding and subtracting the 1.

Thus,

$\int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}} d x = \int \frac{1}{\sqrt{9 x^{2} + 6 x + 1 - 1 - 8}} d x \int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}} d x = \int \frac{1}{\sqrt{{(3 x)}^{2} + 2 (3) (1) x + 1^{2} - 1 - 8}} d x \int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}} d x = \int \frac{1}{\sqrt{{(3 x)}^{2} + 2 (3) (1) x + 1^{2} - 9}} d x \int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}} d x = \int \frac{1}{\sqrt{{(3 x + 1)}^{2} - 9}} d x$

Step 2

Now, consider $3 x + 1 = u . . . . (1)$

Then,

$3 d x = d u d x = \frac{d u}{3} . . . . (2)$

From equations (1) and (2) substitute these values in $\int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}} d x = \int \frac{1}{\sqrt{{(3 x + 1)}^{2} - 9}} d x$ .

$\int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}} d x = \int \frac{1}{\sqrt{u^{2} - 9}} \frac{d u}{3} \int \frac{1}{\sqrt{9 x^{2} + 6 x - 8}} d x = \frac{1}{3} \int \frac{1}{\sqrt{u^{2} - 3^{2}}} d u$