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

The integral of log_{e^x}(x) is (ln^2(x))/2+C

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integral of log_{e^x}(x)

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

\frac{ln ^{2} ( x )}{2} + C

Solution steps

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

Simplify

lo g_{e^{x}} (x) : \frac{ln ( x )}{x}

= \int \frac{ln ( x )}{x} d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{u ^{2}}{2}

Substitute back

u = ln (x)

= \frac{ln ^{2} ( x )}{2}

Add a constant to the solution

= \frac{ln ^{2} ( x )}{2} + C

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What is the integral of log_{e^x}(x) ?
The integral of log_{e^x}(x) is (ln^2(x))/2+C