Explanation Given information: ∫ ln ( x + x 2 ) d x Formula: ∫ u v d x = u ∫ v d x − ∫ d u d x ( ∫ v d x ) d x d f ( u ) d x = d f d u ⋅ d u d x ∫ x a d x = x a + 1 a + 1 ∫ 1 u d u = ln ( | u | ) Calculation: Evaluate the integral. ∫ ln ( x + x 2 ) d x = ln ( x + x 2 ) ∫ ( 1 ) d x − ∫ d ( ln ( x + x 2 ) ) d x ( ∫ ( 1 ) d x ) d x = ln ( x + x 2 ) ( x ) − ∫ d ( ln ( x + x 2 ) ) d x ( x ) d x + C = x ln ( x + x 2 ) − ∫ d ( ln ( x + x 2 ) ) d x ( x ) d x + C Substitute u for ( x + x 2 ) . ∫ ln ( x + x 2 ) d x = x ln ( u ) − ∫ d ( ln ( u ) ) d x ( x ) d x + C Substitute f for ln ( u ) . ∫ ln ( x + x 2 ) d x = x ( f ) − ∫ d ( f ) d x ( x ) d x + C Apply the chain rule. ∫ ln ( x + x 2 ) d x = x ( f ) − ∫ ( d ( f ) d u ⋅ d u d x ) ( x ) d x + C Substitute ln ( u ) for f and ( x + x 2 ) for u. ∫ ln ( x + x 2 ) d x = x ln ( u ) − ∫ ( d ( ln ( u ) ) d u ⋅ d ( x + x 2 ) d x ) ( x ) d x + C = x ln ( u ) − ∫ ( 1 u ( 1 + 2 x ) ) ( x ) d x + C Substitute ( x + x 2 ) for u

University Calculus: Early Transcendentals Plus MyLab Math -- Access Card Package (3rd Edition) (Integrated Review Courses in MyMathLab and MyStatLab)

3rd Edition

ISBN: 9780321999573

Author: Joel R. Hass, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 8.1, Problem 28E

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The integral by using a substitution prior to integration by parts.

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Chapter 8.1, Problem 27E

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