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

The integral of 1/x (1+ln(x)) is ((1+ln(x))^2)/2+C

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

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

\frac{( 1 + ln ( x ) ) ^{2}}{2} + C

Solution steps

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

Simplify

\frac{1}{x} (1 + ln (x)) : \frac{1 + ln ( x )}{x}

= \int \frac{1 + 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 = 1 + ln (x)

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

Add a constant to the solution

= \frac{( 1 + ln ( x ) ) ^{2}}{2} + C

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What is the integral of 1/x (1+ln(x)) ?
The integral of 1/x (1+ln(x)) is ((1+ln(x))^2)/2+C