Take the integral: *x+1 x-1 dx

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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I am stuck on this integral. Can I see the steps for the long division? Is there any other way to do this?

Take the integral:
dx
For the integrand
= √₂/2²/₁ +
x-1
X + 1
X-1
+1 dx
Integrate the sum term by term and factor out constants:
= 2√===dx+f1dx
1
For the integrand
X-1
=2f=du + frax
Answer:
do long division:
, substitute u = x - 1 and du = dx:
3
1
The integral of - is log(u):
U
= 2log(u) + f1dx
The integral of 1 is x:
= 2log(u) + x + constant
Substitute back for u = x - 1:
= x + 2log(x-1)+constant
Transcribed Image Text:Take the integral: dx For the integrand = √₂/2²/₁ + x-1 X + 1 X-1 +1 dx Integrate the sum term by term and factor out constants: = 2√===dx+f1dx 1 For the integrand X-1 =2f=du + frax Answer: do long division: , substitute u = x - 1 and du = dx: 3 1 The integral of - is log(u): U = 2log(u) + f1dx The integral of 1 is x: = 2log(u) + x + constant Substitute back for u = x - 1: = x + 2log(x-1)+constant
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