What is the integral of e^{-x}arccot(e^x) ?

The integral of e^{-x}arccot(e^x) is -e^{-x}arccot(e^x)+1/2 ln|e^{2x}+1|-x+C

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integral of e^{-x}arccot(e^x)

\int e^{- x} arccot (e^{x}) d x

- e^{- x} arccot (e^{x}) + \frac{1}{2} ln e^{2 x} + 1 - x + C

Solution steps

\int e^{- x} arccot (e^{x}) d x

Apply Integration By Parts

: - e^{- x} arccot (e^{x}) - \int \frac{1}{e ^{2 x} + 1} d x

= - e^{- x} arccot (e^{x}) - \int \frac{1}{e ^{2 x} + 1} d x

\int \frac{1}{e ^{2 x} + 1} d x = - \frac{1}{2} ln e^{2 x} + 1 + x

= - e^{- x} arccot (e^{x}) - (- \frac{1}{2} ln e^{2 x} + 1 + x)

Simplify

= - e^{- x} arccot (e^{x}) + \frac{1}{2} ln e^{2 x} + 1 - x

Add a constant to the solution

= - e^{- x} arccot (e^{x}) + \frac{1}{2} ln e^{2 x} + 1 - x + C

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What is the integral of e^{-x}arccot(e^x) ?
The integral of e^{-x}arccot(e^x) is -e^{-x}arccot(e^x)+1/2 ln|e^{2x}+1|-x+C