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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Using Integration Tables In Exercises 9–36, use the integration table in Appendix C to find the indefinite integral. See Examples 1, 2,3, and 5.

1 x 2 x 2 4 d x

To determine

To calculate: The solution of indefinite integral 1x2x24dx.

Explanation

Given Information:

The expression is provided as, 1x2x24dx.

Formula used:

The formula for integral 1u2u2±a2du is,

1u2u2±a2du=u2±a2a2u+C

Calculation:

Let u=x.

Differentiate above integral with respect to x,

du=dx

Consider the provided expression,

1x2x24dx

Rewrite.

1x2x24dx=1x2x222dx

Here, a=2, u=x and du=dx.

Substitute u for x, a for 2, and du for dx

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