What is the integral of (arctan(e^x))/(e^x) ?

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

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integral of (arctan(e^x))/(e^x)

\int \frac{arctan ( e ^{x} )}{e ^{x}} d x

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

Solution steps

\int \frac{arctan ( e ^{x} )}{e ^{x}} d x

Apply u-substitution

: \int \frac{arctan ( u )}{u ^{2}} d u

= \int \frac{arctan ( u )}{u ^{2}} d u

Apply Integration By Parts

: - \frac{arctan ( u )}{u} - \int - \frac{1}{u ( u ^{2} + 1 )} d u

= - \frac{arctan ( u )}{u} - \int - \frac{1}{u ( u ^{2} + 1 )} d u

\int - \frac{1}{u ( u ^{2} + 1 )} d u = - ln ∣ u ∣ + \frac{1}{2} ln u^{2} + 1

= - \frac{arctan ( u )}{u} - (- ln ∣ u ∣ + \frac{1}{2} ln u^{2} + 1)

Substitute back

u = e^{x}

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

Simplify

- \frac{arctan ( e ^{x} )}{e ^{x}} - (- ln ∣ e^{x} ∣ + \frac{1}{2} ln (e^{x})^{2} + 1) : - \frac{arctan ( e ^{x} )}{e ^{x}} + x - \frac{1}{2} ln e^{2 x} + 1

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

Add a constant to the solution

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

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x \to 1 lim (\frac{x + 4}{1 + x ^{2}})

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x \to 0 lim (\frac{( a ^{x} - b ^{x} )}{x})

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