What is the integral of (ln(3x+1))/(6x+2) ?

The integral of (ln(3x+1))/(6x+2) is 1/12 ln^2(3x+1)+C

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integral of (ln(3x+1))/(6x+2)

\int \frac{ln ( 3 x + 1 )}{6 x + 2} d x

\frac{1}{12} ln^{2} (3 x + 1) + C

Solution steps

\int \frac{ln ( 3 x + 1 )}{6 x + 2} d x

\frac{ln ( 3 x + 1 )}{6 x + 2} = \frac{1}{2} \frac{ln ( 3 x + 1 )}{3 x + 1}

= \int \frac{1}{2} \cdot \frac{ln ( 3 x + 1 )}{3 x + 1} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{ln ( 3 x + 1 )}{3 x + 1} d x

Apply u-substitution

: \int \frac{ln ( u )}{3 u} d u

= \frac{1}{2} \cdot \int \frac{ln ( u )}{3 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int \frac{ln ( u )}{u} d u

Apply u-substitution

: \int v d v

= \frac{1}{2} \cdot \frac{1}{3} \cdot \int v d v

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{v ^{2}}{2}

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{ln ^{2} ( 3 x + 1 )}{2} : \frac{1}{12} ln^{2} (3 x + 1)

= \frac{1}{12} ln^{2} (3 x + 1)

Add a constant to the solution

= \frac{1}{12} ln^{2} (3 x + 1) + C

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What is the integral of (ln(3x+1))/(6x+2) ?
The integral of (ln(3x+1))/(6x+2) is 1/12 ln^2(3x+1)+C