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

options
  1. π/4e2
  2. π/4e
  3. 1/e2(π/2 – tan-1 1/e)
  4. π/2e2

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Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 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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