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

y^{^{'}} = y (3 - t y)

\int \frac{3 x ^{2} - x + 5}{x ^{4}} d x

y^{^{'}} = - \frac{x}{y}

x \to - 3 lim (\frac{x + 2}{x - 2})

\frac{\partial}{\partial x} (cos^{2} (x) y)

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