EXAMPLE 5 Find Vx x' dx as x4x. Let u 1+x2.Then du 2x dx, so dx = 1du. Also x2 = u - 1, so x4 (u - 1)2: SOLUTION An appropriate substitution becomes more obvious if we factor x = =SVu (u - 1. du= } f Vu (u° _ 2u + 1) du = (u2 - 2u3/2 u2)du =u - 2.Zu5e +zu3n) + C_ =t(1 x2)2 -(1 + x2)%2 + $(1 + x?)2 + C
EXAMPLE 5 Find Vx x' dx as x4x. Let u 1+x2.Then du 2x dx, so dx = 1du. Also x2 = u - 1, so x4 (u - 1)2: SOLUTION An appropriate substitution becomes more obvious if we factor x = =SVu (u - 1. du= } f Vu (u° _ 2u + 1) du = (u2 - 2u3/2 u2)du =u - 2.Zu5e +zu3n) + C_ =t(1 x2)2 -(1 + x2)%2 + $(1 + x?)2 + C
EXAMPLE 5 Find Vx x' dx as x4x. Let u 1+x2.Then du 2x dx, so dx = 1du. Also x2 = u - 1, so x4 (u - 1)2: SOLUTION An appropriate substitution becomes more obvious if we factor x = =SVu (u - 1. du= } f Vu (u° _ 2u + 1) du = (u2 - 2u3/2 u2)du =u - 2.Zu5e +zu3n) + C_ =t(1 x2)2 -(1 + x2)%2 + $(1 + x?)2 + C
I am trying to understand the factoring in the explicit solution of the attached integration problem.
As you can see, they made dx= 1/2 du.
When the x^5 is factored into x^4 and x, they rewrite x^4 as (u-1)^2, but don't explain where or how they factored the x. If dx were equal to 1/2x du I think it would make sense, but as it is, I don't understand where the x went after the factoring.
Transcribed Image Text:EXAMPLE 5 Find Vx x' dx
as x4x. Let u 1+x2.Then du
2x dx, so dx = 1du. Also x2 = u - 1, so x4
(u - 1)2:
SOLUTION An appropriate substitution becomes more obvious if we factor x
=
=SVu (u - 1. du= } f Vu (u° _ 2u + 1) du
= (u2 - 2u3/2 u2)du
=u
- 2.Zu5e +zu3n) + C_
=t(1 x2)2 -(1 + x2)%2 + $(1 + x?)2 + C
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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