What is the integral of e^{2t}arctan(e^t) ?

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

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integral of e^{2t}arctan(e^t)

\int e^{2 t} arctan (e^{t}) d t

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

Solution steps

\int e^{2 t} arctan (e^{t}) d t

Apply u-substitution

: \int u arctan (u) d u

= \int u arctan (u) d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = e^{t}

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

Add a constant to the solution

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

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