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

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

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

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

2 e^{x - 1} + C

Solution steps

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

\frac{e ^{x - 1}}{x} = e^{- 1} \frac{e ^{x}}{x}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e^{- 1} \cdot \int \frac{e ^{x}}{x} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{e} \cdot \int \frac{e ^{x}}{x} d x

Apply u-substitution

: \int 2 e^{u} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{e} \cdot 2 e^{u}

Substitute back

u = x

= \frac{1}{e} \cdot 2 e^{x}

Simplify

\frac{1}{e} \cdot 2 e^{x} : 2 e^{x - 1}

= 2 e^{x - 1}

Add a constant to the solution

= 2 e^{x - 1} + C

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