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∫ dx/e x +4e -x = f (x) + c then f (x) is equal to options 1)2 tan -1 (2e x ) 1/2 tan-1 (ex/2) 2tan -1 e x /2 1/2 tan -1 (2e 2x )

∫ dx/ex +4e-x =f (x) + c then f (x) is equal to
options
1)2 tan-1 (2ex)
  1.  1/2 tan-1 (ex/2)
  2. 2tan-1 ex/2
  3. 1/2 tan-1 (2e2x)

Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question below

\int \frac{dx}{e^{x} + 4e^{-x}} = f(x) + c
I = \int \frac{dx}{e^{x} + 4e^{-x}}
I = \int \frac{e^{x}}{e^{2x}+4}dx
I = \int \frac{e^{x}}{(e^{x})^{2}+(2)^{2}}dx
e^{x} = t
e^{x}dx = dt
I = \int \frac{1}{t^{2}+(2)^{2}}dt
I = \frac{1}{2}tan^{-1}(\frac{t}{2}) + c
I = \frac{1}{2}tan^{-1}(\frac{e^{x}}{2}) + c


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