What is the integral of 1/(7^{1/x)} ?

The integral of 1/(7^{1/x)} is 7^{-1/x}x+ln(7)\Ei(-ln(7^{1/x}))+C

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integral of 1/(7^{1/x)}

integral

\frac{1}{7 ^{\frac{1}{x}}}

7^{- \frac{1}{x}} x + ln (7) Ei (- ln (7^{\frac{1}{x}})) + C

Solution steps

\int \frac{1}{7 ^{\frac{1}{x}}} d x

Apply u-substitution

: \int - \frac{ln ( 7 )}{u ^{2} ln ^{2} ( u )} d u

= \int - \frac{ln ( 7 )}{u ^{2} ln ^{2} ( u )} d u

Take the constant out:

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

= - ln (7) \cdot \int \frac{1}{u ^{2} ln ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{e ^{v} v ^{2}} d v

= - ln (7) \cdot \int \frac{1}{e ^{v} v ^{2}} d v

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{v} v ^{2}} = e^{- v} \frac{1}{v ^{2}}

= - ln (7) \cdot \int e^{- v} \frac{1}{v ^{2}} d v

Simplify

e^{- v} \frac{1}{v ^{2}} : \frac{e ^{- v}}{v ^{2}}

= - ln (7) \cdot \int \frac{e ^{- v}}{v ^{2}} d v

Apply Integration By Parts

: - \frac{e ^{- v}}{v} - \int \frac{e ^{- v}}{v} d v

= - ln (7) (- \frac{e ^{- v}}{v} - \int \frac{e ^{- v}}{v} d v)

\int \frac{e ^{- v}}{v} d v = Ei (- v)

= - ln (7) (- \frac{e ^{- v}}{v} - Ei (- v))

Substitute back

= - ln (7) - \frac{e ^{- l n (7^{\frac{1}{x}})}}{ln ( 7 ^{\frac{1}{x}} )} - Ei (- ln (7^{\frac{1}{x}}))

Simplify

- ln (7) - \frac{e ^{- l n (7^{\frac{1}{x}})}}{ln ( 7 ^{\frac{1}{x}} )} - Ei (- ln (7^{\frac{1}{x}})) : 7^{- \frac{1}{x}} x + ln (7) Ei (- ln (7^{\frac{1}{x}}))

= 7^{- \frac{1}{x}} x + ln (7) Ei (- ln (7^{\frac{1}{x}}))

Add a constant to the solution

= 7^{- \frac{1}{x}} x + ln (7) Ei (- ln (7^{\frac{1}{x}})) + C

\frac{sin ( x )}{x ^{2}}

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

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

cos (\frac{x}{2})

\frac{x ^{3}}{( x + 1 ) ^{2}}

What is the integral of 1/(7^{1/x)} ?
The integral of 1/(7^{1/x)} is 7^{-1/x}x+ln(7)\Ei(-ln(7^{1/x}))+C