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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

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

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

= 2 (e^{u} u - e^{u})

Substitute back

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

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

Add a constant to the solution

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

y = (4 x + 3)^{5}

x^{- \frac{1}{3}}

f (x) = x - (x + 6)^{6}

x \to 0 lim (\frac{ln ( 1 + 10 x )}{x})

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