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

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

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integral of 1/(sqrt(e^{2x-1))}

\int \frac{1}{e ^{2 x - 1}} d x

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

Solution steps

\int \frac{1}{e ^{2 x - 1}} d x

Apply u-substitution

: \int \frac{1}{2 e ^{\frac{u}{2}}} d u

= \int \frac{1}{2 e ^{\frac{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{1}{e ^{\frac{u}{2}}} d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = 2 x - 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{y ^{2}}{x})

\int \frac{7 x ^{3} - 7 x + 5}{x ^{3} - x ^{2}} d x

\frac{d}{d x} (\frac{20}{x})

x \to m + lim (x - [x])

\frac{d}{d x} (arctan (e^{2 x}))

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