What is the integral of (2ln(x)-1)/x ?

The integral of (2ln(x)-1)/x is 1/4 (2ln(x)-1)^2+C

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integral of (2ln(x)-1)/x

\int \frac{2 ln ( x ) - 1}{x} d x

\frac{1}{4} (2 ln (x) - 1)^{2} + C

Solution steps

\int \frac{2 ln ( x ) - 1}{x} d x

Apply u-substitution

: \int \frac{u}{2} d u

= \int \frac{u}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{u ^{2}}{2}

Substitute back

u = 2 ln (x) - 1

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

Simplify

\frac{1}{2} \cdot \frac{( 2 ln ( x ) - 1 ) ^{2}}{2} : \frac{1}{4} (2 ln (x) - 1)^{2}

= \frac{1}{4} (2 ln (x) - 1)^{2}

Add a constant to the solution

= \frac{1}{4} (2 ln (x) - 1)^{2} + C

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What is the integral of (2ln(x)-1)/x ?
The integral of (2ln(x)-1)/x is 1/4 (2ln(x)-1)^2+C