What is the integral of log_{e^1}(x) ?

The integral of log_{e^1}(x) is xln(x)-x+C

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integral of log_{e^1}(x)

\int lo g_{e^{1}} (x) d x

x ln (x) - x + C

Solution steps

\int lo g_{e^{1}} (x) d x

Simplify

lo g_{e^{1}} (x) : ln (x)

= \int ln (x) d x

Apply Integration By Parts

: x ln (x) - \int 1 d x

= x ln (x) - \int 1 d x

\int 1 d x = x

= x ln (x) - x

Add a constant to the solution

= x ln (x) - x + C

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What is the integral of log_{e^1}(x) ?
The integral of log_{e^1}(x) is xln(x)-x+C