Integrate each of the following functions using substitution, finding the most general antiderivative. Also enter u, the function of x that you substitute. If your answer requires a constant of integration, enter it as c. ❤ b. 5x4 u = 2/3(x^5+1)^(3/2)+c 4x³ +3 J √x²+3x+4 U = J e. Vào U= J U= J (In x)4 U = X +1 dx U= a J x¹ (2x³ + 2)6 dx dx X 5x5 +2 x² 6x³ +5 dx dx = dx = 2/3(x^5+1)^(3/2)+c = 1/5ln^5(x)+c = 2(x^4+3x+4)^(1/2)+c = 1/70(2x^5+2)^7+c 1/9(6x^3 +5)^(1/2)+c = 1/25In(abs(5x^5+2))+c ✪ O Ö

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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q-f only the parts asking for U=

Integrate each of the following functions using substitution, finding the most general antiderivative.
Also enter u, the function of x that you substitute.
If your answer requires a constant of integration, enter it as c.
a.
u = 2/3(x^5+1)^(3/2)+c
4x³ +3
J
x4 + 3x + 4
b.
5x²√x5 + 1 dx
52
d.
U =
U=
(In x) ¹
U=
J
u =
4
dx
U=
x²
S
Vor +5 dr
e.
5x5 +2
=
(2x³ + 2)6 dx
dx
dx
2/3(x^5+1)^(3/2)+c
=
1/5ln^5(x)+c
|2(x^4+3x+4)^(1/2)+c
=
1/70(2x^5+2)^7+c
1/9(6x^3 +5)^(1/2)+c
= 1/25In(abs(5x^5+2))+c
X
Transcribed Image Text:Integrate each of the following functions using substitution, finding the most general antiderivative. Also enter u, the function of x that you substitute. If your answer requires a constant of integration, enter it as c. a. u = 2/3(x^5+1)^(3/2)+c 4x³ +3 J x4 + 3x + 4 b. 5x²√x5 + 1 dx 52 d. U = U= (In x) ¹ U= J u = 4 dx U= x² S Vor +5 dr e. 5x5 +2 = (2x³ + 2)6 dx dx dx 2/3(x^5+1)^(3/2)+c = 1/5ln^5(x)+c |2(x^4+3x+4)^(1/2)+c = 1/70(2x^5+2)^7+c 1/9(6x^3 +5)^(1/2)+c = 1/25In(abs(5x^5+2))+c X
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