The correct option is B tan ^ 1e^x. Let nbsp;I=∫ nbsp; 1 / e^ x +e^ x dx=∫ nbsp; e^ x dx / e^ 2x +1 dx Put nbsp; e ^ x =t⇒ e ^ x dx=dt ...

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Functions

∫1/ex+e-xdx.

Question

$\int \frac{1}{e^{x} + e^{- x}} d x .$

{tan}^{- 1} x .

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{tan}^{- 1} e^{x} .

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{cot}^{- 1} e^{x} .

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\frac{1}{2} {cot}^{- 1} e^{x} .

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Solution

The correct option is B ${tan}^{- 1} e^{x} .$
Let $I = \int \frac{1}{e^{x} + e^{- x}} d x = \int \frac{e^{x} d x}{e^{2 x} + 1} d x$
Put $e^{x} = t \Rightarrow e^{x} d x = d t$
$I = \int \frac{d t}{t^{2} + 1} = {tan}^{- 1} t = {tan}^{- 1} e^{x}$

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