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

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

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

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

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

Solution steps

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

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^{x}

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

Apply exponent rule:

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

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

Add a constant to the solution

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

arccos (5 x)

\frac{d}{d x} (x^{4} (x - 2)^{3})

y^{^{''}} - 4 y^{^{'}} + 3 y = 7 x^{3} e^{(2 x)} cos (2 x)

x \to 0 lim (\frac{1}{e ^{x} + 1})

\int \frac{sin ^{3} ( x )}{cos ^{7} ( x )} d x

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