What is the integral of 1/(1-ln(x)) ?

The integral of 1/(1-ln(x)) is -e\Ei(-1+ln(x))+C

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integral of 1/(1-ln(x))

\int \frac{1}{1 - ln ( x )} d x

- e Ei (- 1 + ln (x)) + C

Solution steps

\int \frac{1}{1 - ln ( x )} d x

Factor

1 - ln (x) : - (- 1 + ln (x))

= \int \frac{1}{- ( - 1 + ln ( x ) )} d x

Take the constant out:

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

= - \int \frac{1}{- 1 + ln ( x )} d x

Apply Integration By Parts

= - (\frac{x}{- 1 + ln ( x )} - \int - \frac{1}{( - 1 + ln ( x ) ) ^{2}} d x)

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

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

Simplify

= - e Ei (- 1 + ln (x))

Add a constant to the solution

= - e Ei (- 1 + ln (x)) + C

What is the integral of 1/(1-ln(x)) ?
The integral of 1/(1-ln(x)) is -e\Ei(-1+ln(x))+C