What is the integral of (ln(ax+b))/(ax+b) ?

The integral of (ln(ax+b))/(ax+b) is 1/(2a)ln^2(ax+b)+C

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integral of (ln(ax+b))/(ax+b)

\int \frac{ln ( a x + b )}{a x + b} d x

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

Solution steps

\int \frac{ln ( a x + b )}{a x + b} d x

Apply u-substitution

: \int \frac{ln ( u )}{a u} d u

= \int \frac{ln ( u )}{a u} d u

Take the constant out:

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

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

Apply u-substitution

: \int v d v

= \frac{1}{a} \cdot \int v d v

Apply the Power Rule

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

= \frac{1}{a} \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

\frac{1}{a} \cdot \frac{ln ^{2} ( a x + b )}{2} : \frac{1}{2 a} ln^{2} (a x + b)

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

Add a constant to the solution

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

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