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

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

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integral of ln(1+3x)

\int ln (1 + 3 x) d x

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

Solution steps

\int ln (1 + 3 x) 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 = 1 + 3 x

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

Simplify

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

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

Add a constant to the solution

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

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