What is the integral of (ln(x))' ?

The integral of (ln(x))' is x(ln(x))'+ln|x|+C

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integral of (ln(x))'

\int (ln (x))^{'} d x

x (ln (x))' + ln ∣ x ∣ + C

Solution steps

\int (ln (x))^{'} d x

Apply Integration By Parts

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

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

\int - \frac{1}{x} d x = - ln ∣ x ∣

= x (ln (x))^{^{'}} - (- ln ∣ x ∣)

Simplify

= x (ln (x))^{^{'}} + ln ∣ x ∣

Add a constant to the solution

= x (ln (x))^{^{'}} + ln ∣ x ∣ + C

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What is the integral of (ln(x))' ?
The integral of (ln(x))' is x(ln(x))'+ln|x|+C