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

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

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

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

- \frac{1}{ln ( 2 )} x e^{- l n (2) \frac{1}{x} t} + C

Solution steps

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

Apply u-substitution

: \int - \frac{e ^{\frac{u}{x}}}{ln ( 2 )} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int e^{v} x d v

= - \frac{1}{ln ( 2 )} \cdot \int e^{v} x d v

Take the constant out:

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

= - \frac{1}{ln ( 2 )} x \cdot \int e^{v} d v

Use the common integral:

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

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

Substitute back

= - \frac{1}{ln ( 2 )} x e^{\frac{- l n ( 2 ) t}{x}}

Simplify

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

= - \frac{1}{ln ( 2 )} x e^{- l n (2) \frac{1}{x} t}

Add a constant to the solution

= - \frac{1}{ln ( 2 )} x e^{- l n (2) \frac{1}{x} t} + C

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