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

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

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

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

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

Solution steps

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

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 x}

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

Simplify

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

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

Add a constant to the solution

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

y = x^{4^{x}}

\int 9 x e^{5 x^{2}} d x

\frac{d}{d x} (A (e^{x} + e^{x} x))

x^{2} + x y + 2 y^{2} = 4, (- 2, 1)

9 y^{^{''}} + 6 y^{^{'}} + y = 0, y (- 3) = - 2, y^{^{'}} (1) = - 1

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