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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

x \to 1 lim (3 x - 6)

\frac{2 s e ^{- 2 s}}{5 s ^{2} + 45}

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

y^{^{''}} - y^{^{'}} - 2 y = 4 e^{- t}

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