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

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

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

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

2 e^{x^{\frac{1}{2}}} + C

Solution steps

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

Apply u-substitution

: \int 2 e^{u} d u

= \int 2 e^{u} d u

Take the constant out:

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

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

Use the common integral:

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

= 2 e^{u}

Substitute back

u = x^{\frac{1}{2}}

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

Add a constant to the solution

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

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