What is the integral of ln(3x-2) ?

The integral of ln(3x-2) is xln(3x-2)-2/3 ln(3x-2)-x+2/3+C

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integral of ln(3x-2)

integral

ln (3 x - 2)

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

Solution steps

\int ln (3 x - 2) d x

Apply u-substitution

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

Take the constant out:

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

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

Apply Integration By Parts

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

\int 1 d u = u

= \frac{1}{3} (u ln (u) - u)

Substitute back

u = 3 x - 2

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

Simplify

\frac{1}{3} ((3 x - 2) ln (3 x - 2) - (3 x - 2)) : x ln (3 x - 2) - \frac{2}{3} ln (3 x - 2) - x + \frac{2}{3}

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

Add a constant to the solution

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

in t e g r a l \frac{( 1 + 2 x )}{( 1 + x )}

in t e g r a l \frac{9}{x ^{5}}

in t e g r a l 3 x + 2

in t e g r a l \frac{( y + 1 ) ^{2}}{y}

in t e g r a l (1 - y)^{4}

What is the integral of ln(3x-2) ?
The integral of ln(3x-2) is xln(3x-2)-2/3 ln(3x-2)-x+2/3+C