What is the integral of (e^{6x})/(1+e^{2x)} ?

The integral of (e^{6x})/(1+e^{2x)} is 1/4 (e^{4x}+2ln|e^{2x}+1|-2e^{2x}-3)+C

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integral of (e^{6x})/(1+e^{2x)}

\int \frac{e ^{6 x}}{1 + e ^{2 x}} d x

\frac{1}{4} (e^{4 x} + 2 ln e^{2 x} + 1 - 2 e^{2 x} - 3) + C

Solution steps

\int \frac{e ^{6 x}}{1 + e ^{2 x}} d x

Apply u-substitution

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

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = 1 + e^{2 x}

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

Simplify

\frac{1}{2} (\frac{( 1 + e ^{2 x} ) ^{2}}{2} - 2 (1 + e^{2 x}) + ln 1 + e^{2 x}) : \frac{1}{4} (e^{4 x} + 2 ln e^{2 x} + 1 - 2 e^{2 x} - 3)

= \frac{1}{4} (e^{4 x} + 2 ln e^{2 x} + 1 - 2 e^{2 x} - 3)

Add a constant to the solution

= \frac{1}{4} (e^{4 x} + 2 ln e^{2 x} + 1 - 2 e^{2 x} - 3) + C

\frac{d}{d x} (\frac{x ( c ^{2} - x ^{2} )}{2.4 ( c ^{2} + c x )})

\frac{d y}{d x} = y (2 - y)

\int 2 x (x^{2} + 19)^{14} d x

\int (x^{2} - 5)^{5} 2 x d x

\int 7 arctan (x) d x

What is the integral of (e^{6x})/(1+e^{2x)} ?
The integral of (e^{6x})/(1+e^{2x)} is 1/4 (e^{4x}+2ln|e^{2x}+1|-2e^{2x}-3)+C