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

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

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

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

- \frac{a ^{\frac{1}{x}}}{ln ( a )} + C

Solution steps

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

Apply u-substitution

: \int - a^{u} d u

= \int - a^{u} d u

Take the constant out:

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

= - \int a^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - \frac{a ^{u}}{ln ( a )}

Substitute back

u = \frac{1}{x}

= - \frac{a ^{\frac{1}{x}}}{ln ( a )}

Add a constant to the solution

= - \frac{a ^{\frac{1}{x}}}{ln ( a )} + C

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