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Integral Calculus> options π/4e 2 π/4e 1/e 2 (π/2 – tan -1 1...

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options π/4e2 π/4e 1/e2(π/2 – tan-1 1/e) π/2e2

Aditya Kartikeya , 10 Years ago

Grade 10

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$I = \int_{1}^{\infty }\frac{1}{e^{x+1}+e^{3-x}}dx$
$I = \int_{1}^{\infty }\frac{e^{x}}{e^{2x+1}+e^{3}}dx$
$I = \frac{1}{e}\int_{1}^{\infty }\frac{e^{x}}{e^{2x}+e^{2}}dx$
$I = \frac{1}{e}\int_{1}^{\infty }\frac{e^{x}}{(e^{x})^{2}+e^{2}}dx$
$e^{x} = t$
$e^{x}dx = dt$
$I = \frac{1}{e}\int_{e}^{\infty }\frac{1}{t^2+e^2}dt$
$I = \frac{1}{e}.[\frac{1}{e}.tan^{-1}(\frac{t}{e})]_{e}^{\infty}$
$I = \frac{1}{e^2}.[\frac{\pi }{2} - \frac{\pi }{4}]$
$I = \frac{1}{e^2}.[\frac{\pi }{4}]$
$I = \frac{\pi }{4e^2}$
Option (1) is correct.

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