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Evaluate the integral of arccot ex / ex

by RoRi

Evaluate the following integral

    \[ \int \frac{\operatorname{arccot} e^x}{e^x} \, dx. \]

First, let us make the substitution s = e^x which implies ds = e^x \, dx (which further implies that dx = \frac{ds}{s}). Then we have

    \[ \int \frac{\operatorname{arccot} e^x}{e^x} \, dx &= \int \frac{\operatorname{arccot} s}{s^2} \, ds. \]

Next, we integrate by parts with

    \begin{align*} u &= \operatorname{arccot} s & du &= \frac{-1}{1+s^2} \, dx \\ dv &= \frac{1}{s^2} \, ds & v &= \frac{-1}{s}. \end{align*}

Therefore, we have

    \begin{align*} \int \frac{\operatorname{arccot} s}{s^2} \, ds &= -\frac{\operatorname{arccot} s}{s} - \int \frac{1}{s+s^3} \,ds \\[9pt] &= -\frac{\operatorname{arccot} s}{s} - \int \frac{1+s^2-s^2}{s(1+s^2)} \, ds \\[9pt] &= -\frac{\operatorname{arccot} s}{s} - \int \frac{1}{s} \, ds + \int \frac{s}{1+s^2} \, ds \\[9pt] &= -\frac{\operatorname{arccot} s}{s} - \log |s| + \frac{1}{2} \log(1+s^2) + C \\[9pt] &= -\frac{\operatorname{arccot} e^x}{e^x} + \frac{1}{2} \log \frac{1+s^2}{s^2} + C\\[9pt] &= \frac{1}{2} \log (1+e^{-2x}) - \frac{\operatorname{arccot} e^x}{e^x} + C. \end{align*}

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