What is the integral of e^ysqrt(1+e^{2y)} ?

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

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integral of e^ysqrt(1+e^{2y)}

\int e^{y} 1 + e^{2 y} d y

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

Solution steps

\int e^{y} 1 + e^{2 y} d y

Apply u-substitution

: \int \frac{1 + u}{2 u} d u

= \int \frac{1 + u}{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{1 + u}{u} d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = e^{2 y}

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

Simplify

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

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

Add a constant to the solution

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

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