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

The integral of ln(3x+1) is xln(3x+1)+1/3 ln(3x+1)-x-1/3+C

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integral of ln(3x+1)

\int ln (3 x + 1) d x

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

Solution steps

\int ln (3 x + 1) d x

Apply u-substitution

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

= \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

: u ln (u) - \int 1 d u

= \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 + 1

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

Simplify

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

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

Add a constant to the solution

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

f (x) = x^{3} - 7 x^{2}

\frac{d y}{d x} = (e^{x} + 10)^{2}

\frac{\partial}{\partial x} (\frac{( x - y )}{z})

\int (sin (x) e^{2 x}) d x

\frac{d}{d x} (sin (\frac{5}{x}))

What is the integral of ln(3x+1) ?
The integral of ln(3x+1) is xln(3x+1)+1/3 ln(3x+1)-x-1/3+C