What is the integral of (e^{3y})/(1+e^y) ?

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

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integral of (e^{3y})/(1+e^y)

\int \frac{e ^{3 y}}{1 + e ^{y}} d y

\frac{( 1 + e ^{y} ) ^{2}}{2} - 2 (1 + e^{y}) + ln ∣ 1 + e^{y} ∣ + C

Solution steps

\int \frac{e ^{3 y}}{1 + e ^{y}} d y

Apply u-substitution

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

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

Expand

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

= \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

= \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{u ^{2}}{2} - 2 u + ln ∣ u ∣

Substitute back

u = 1 + e^{y}

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

Add a constant to the solution

= \frac{( 1 + e ^{y} ) ^{2}}{2} - 2 (1 + e^{y}) + ln ∣ 1 + e^{y} ∣ + C

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