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).
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
Problem 1RQ
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