Solue18 (6) and 18(d)18.OX Part d you may oassume. thal a?+bx²+\ dx= 2T18b.18d.abdx2(atb)IT(x²+ a*) (x? tb*J

Name: Solue18 (6) and 18(d)18.OX Part d you may oassume. thal a?+bx²+\ dx=
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Description: Solue18 (6) and 18(d)18.OX Part d you may oassume. thal a?+bx²+\ dx= 2T18b.18d.abdx2(atb)IT(x²+ a*) (x? tb*J

Here 18b ∫-∞∞x2+1x4+1dx=2π18d ∫0∞abx2+a2x2+b2dx=π2a+b

Answered: Solue 18 Cb) and 18(d) Tox Part d you…

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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Solue
18 (6) and 18(d)
18.
OX Part d you may oassume. thal a?+b
x²+\ dx= 2T
18b.
18d.
ab
dx
2(atb)
IT
(x²+ a*) (x? tb*J

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$18 (b) \int_{- \infty}^{\infty} \frac{x^{2} + 1}{x^{4} + 1} d x = \sqrt{2} π 18 (d) \int_{0}^{\infty} \frac{a b}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x = \frac{π}{2 (a + b)}$

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