Find the exact value of the definite integral below, using the substitution -1/2 X U = x- Let u = x - du dx = 1+ When x When x = 1, u = 0 = (√6+√2) 1 x² + 1 1+x² 1+x4 Let m = x² 1 =(√6 + √2), u = ²/(√6 + √2) - (√6+ √2). dx = √2 ļ 0 √2 0 (√6+√2) = 1 √2 :! 0 1+x² x² 1+x4 x² 1+x4 √6-√2 U= = 1/(√6 + √2)- (6-2), u =(√6 + √2)(√6-√2) = √2 X 1+x² 1+x4 du m 1+ m² x² + 1 du dx = [/loge (1 + m²)], ² du - -(1+1)-(1)-1,3 +2).

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Find the exact value of the definite integral below, using the substitution
1
u = x - ²/₁
X
1
Let u = x=
du
dx
= 1+
(ZA+9^)/²/2
X
1
1
x² +1
x²
1+x²
1+x4
Let m = x²
When x = 1, u = 0
1
-
When x = 1/(√6 + √2), u = /²/ (√6 + √2) — 2²/(√6 + √²),
dx =
||
=
√2
0
√2
0
√2
(Z^+9/^)/²/
0
1
n
u = 1/(√6 + √2)
2
1+x²
1+x4
1+x²
x²
X
1+x4 × x² +1
x²
1+x4
du
=(√6 + √2)(√6-√2) = √2
of nouonut air
m
1+ m²
xp -
du
-
=
- Flog.(1 + m²)]²
du
√6-√2
(6-2),
Aont sdayribu's
1
=log (1+2) -log(1) = loge3
Transcribed Image Text:Find the exact value of the definite integral below, using the substitution 1 u = x - ²/₁ X 1 Let u = x= du dx = 1+ (ZA+9^)/²/2 X 1 1 x² +1 x² 1+x² 1+x4 Let m = x² When x = 1, u = 0 1 - When x = 1/(√6 + √2), u = /²/ (√6 + √2) — 2²/(√6 + √²), dx = || = √2 0 √2 0 √2 (Z^+9/^)/²/ 0 1 n u = 1/(√6 + √2) 2 1+x² 1+x4 1+x² x² X 1+x4 × x² +1 x² 1+x4 du =(√6 + √2)(√6-√2) = √2 of nouonut air m 1+ m² xp - du - = - Flog.(1 + m²)]² du √6-√2 (6-2), Aont sdayribu's 1 =log (1+2) -log(1) = loge3
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