What is the integral of (e^{3x})/((4+e^{2x))^2} ?

The integral of (e^{3x})/((4+e^{2x))^2} is 1/2 (-(e^x}{4+e^{2x}}+1/2 arctan(\frac{e^x)/2))+C

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integral of (e^{3x})/((4+e^{2x))^2}

\int \frac{e ^{3 x}}{( 4 + e ^{2 x} ) ^{2}} d x

\frac{1}{2} (- \frac{e ^{x}}{4 + e ^{2 x}} + \frac{1}{2} arctan (\frac{e ^{x}}{2})) + C

Solution steps

\int \frac{e ^{3 x}}{( 4 + e ^{2 x} ) ^{2}} d x

Apply u-substitution

: \int \frac{u - 4}{2 u ^{2}} d u

= \int \frac{u - 4}{2 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u - 4}{u ^{2}} d u

Apply Integration By Parts

: - \frac{u - 4}{u} - \int - \frac{1}{2 u u - 4} d u

= \frac{1}{2} (- \frac{u - 4}{u} - \int - \frac{1}{2 u u - 4} d u)

\int - \frac{1}{2 u u - 4} d u = - \frac{1}{2} arctan (\frac{1}{2} u - 4)

= \frac{1}{2} (- \frac{u - 4}{u} - (- \frac{1}{2} arctan (\frac{1}{2} u - 4)))

Substitute back

u = 4 + e^{2 x}

= \frac{1}{2} (- \frac{4 + e ^{2 x} - 4}{4 + e ^{2 x}} - (- \frac{1}{2} arctan (\frac{1}{2} 4 + e^{2 x} - 4)))

Simplify

\frac{1}{2} (- \frac{4 + e ^{2 x} - 4}{4 + e ^{2 x}} - (- \frac{1}{2} arctan (\frac{1}{2} 4 + e^{2 x} - 4))) : \frac{1}{2} (- \frac{e ^{x}}{4 + e ^{2 x}} + \frac{1}{2} arctan (\frac{e ^{x}}{2}))

= \frac{1}{2} (- \frac{e ^{x}}{4 + e ^{2 x}} + \frac{1}{2} arctan (\frac{e ^{x}}{2}))

Add a constant to the solution

= \frac{1}{2} (- \frac{e ^{x}}{4 + e ^{2 x}} + \frac{1}{2} arctan (\frac{e ^{x}}{2})) + C

f (x) = \frac{8}{x + 11}, at x = 5

\int_{0}^{1} 3 x d x

x y^{^{'}} = x - y

(x - 1)^{2}, x + 1

y = (t + 3) in (t)

What is the integral of (e^{3x})/((4+e^{2x))^2} ?
The integral of (e^{3x})/((4+e^{2x))^2} is 1/2 (-(e^x}{4+e^{2x}}+1/2 arctan(\frac{e^x)/2))+C