What is the integral of (ln(-2x+3))/(-4x+6) ?

The integral of (ln(-2x+3))/(-4x+6) is -1/8 ln^2(-2x+3)+C

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integral of (ln(-2x+3))/(-4x+6)

\int \frac{ln ( - 2 x + 3 )}{- 4 x + 6} d x

- \frac{1}{8} ln^{2} (- 2 x + 3) + C

Solution steps

\int \frac{ln ( - 2 x + 3 )}{- 4 x + 6} d x

Apply u-substitution

: \int - \frac{ln ( u )}{4 u} d u

= \int - \frac{ln ( u )}{4 u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= - \frac{1}{4} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= - \frac{1}{4} \cdot \frac{v ^{2}}{2}

Substitute back

= - \frac{1}{4} \cdot \frac{ln ^{2} ( - 2 x + 3 )}{2}

Simplify

- \frac{1}{4} \cdot \frac{ln ^{2} ( - 2 x + 3 )}{2} : - \frac{1}{8} ln^{2} (- 2 x + 3)

= - \frac{1}{8} ln^{2} (- 2 x + 3)

Add a constant to the solution

= - \frac{1}{8} ln^{2} (- 2 x + 3) + C

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What is the integral of (ln(-2x+3))/(-4x+6) ?
The integral of (ln(-2x+3))/(-4x+6) is -1/8 ln^2(-2x+3)+C