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

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

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

\int e^{\frac{- 1}{2} l n (x)} d x

2 x + C

Solution steps

\int e^{\frac{- 1}{2} l n (x)} d x

Simplify

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

= \int \frac{1}{x} d x

Apply the Power Rule

: 2 x^{\frac{1}{2}}

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

Apply radical rule:

= 2 x

Add a constant to the solution

= 2 x + C

\int \frac{1}{u ^{2} + 9} d u

\frac{d}{d x} (lo g_{e} (sin (x)))

t (t cos (t) - sin (t))

n = 1 \sum \infty \frac{- 1 ^{n}}{n}

x \to 2 lim (\frac{x ^{2} - 3 x + 2}{x ^{2} + 5 x - 14})

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