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When applying integration by parts to the integral I = [In(h (x))da, with h a differentiable function, you obtain. O a) I= xh' (x) h (x) - Sal x ln(h (x))dx Ob) I = x ln(h (x)) + √2h (a) da xh' (x) dx O c) I = a ln(h (x)) - h (a) -S³ Od) I= ch' (x) + h(x) -fa · [ x ln(h (x))dx

When applying integration by parts to the integral I = [In(h (x))da, with h a differentiable function, you obtain. O a) I= xh' (x) h (x) - Sal x ln(h (x))dx Ob) I = x ln(h (x)) + √2h (a) da xh' (x) dx O c) I = a ln(h (x)) - h (a) -S³ Od) I= ch' (x) + h(x) -fa · [ x ln(h (x))dx

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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When applying integration by parts to the integral 1 = [ln(h (x))da, with h a differentiable function, you obtain. O a) I = ch' (x) h (x) - S x In(h (x)) da Ob) I= x ln(h (x)) + -S³ xh' (x) h (x) -dx xh' (x) ○ c) I = x ln(h (x)) – [ - dx S h (x) xh' (x) + f x ln(h (x))dæ Od) I= h (x)
Transcribed Image Text:When applying integration by parts to the integral 1 = [ln(h (x))da, with h a differentiable function, you obtain. O a) I = ch' (x) h (x) - S x In(h (x)) da Ob) I= x ln(h (x)) + -S³ xh' (x) h (x) -dx xh' (x) ○ c) I = x ln(h (x)) – [ - dx S h (x) xh' (x) + f x ln(h (x))dæ Od) I= h (x)
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