Find the mistake in this working out and fix it. Find the exact value of the definite integral below, using the substitution1u = x - ²/₁X1Let u = x=dudx= 1+(ZA+9^)/²/2X11x² +1x²1+x²1+x4Let m = x²When x = 1, u = 01-When x = 1/(√6 + √2), u = /²/ (√6 + √2) — 2²/(√6 + √²),dx =||=√20√20√2(Z^+9/^)/²/01nu = 1/(√6 + √2)21+x²1+x41+x²x²X1+x4 × x² +1x²1+x4du=(√6 + √2)(√6-√2) = √2of nouonut airm1+ m²xp -du-=- Flog.(1 + m²)]²du√6-√2(6-2),Aont sdayribu's1=log (1+2) -log(1) = loge3

Explanation of the answer is as follows

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

See similar textbooks

Similar questions

send handwritten solution asap
Example 31.7. Solve y,2-4y1+ 3y, = 5".
Find product solution for this PDE: xuy - yuz = 0. %3D
give the answer of last question only dear thanks
The domain of function Ax) = In(2 – x) is %3D
.2 dt t2 + 1
(#2) can someone help, i’m super confused