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

The integral of 2ln(3x+1) is 2xln(3x+1)+2/3 ln(3x+1)-2x-2/3+C

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integral of 2ln(3x+1)

\int 2 ln (3 x + 1) d x

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

Solution steps

\int 2 ln (3 x + 1) d x

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Apply Integration By Parts

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

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

\int 1 d u = u

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

Substitute back

u = 3 x + 1

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 2ln(3x+1) ?
The integral of 2ln(3x+1) is 2xln(3x+1)+2/3 ln(3x+1)-2x-2/3+C