What is the integral of ln(3x+7) ?

The integral of ln(3x+7) is xln(3x+7)+7/3 ln(3x+7)-x-7/3+C

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integral of ln(3x+7)

\int ln (3 x + 7) d x

x ln (3 x + 7) + \frac{7}{3} ln (3 x + 7) - x - \frac{7}{3} + C

Solution steps

\int ln (3 x + 7) 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 + 7

= \frac{1}{3} ((3 x + 7) ln (3 x + 7) - (3 x + 7))

Simplify

\frac{1}{3} ((3 x + 7) ln (3 x + 7) - (3 x + 7)) : x ln (3 x + 7) + \frac{7}{3} ln (3 x + 7) - x - \frac{7}{3}

= x ln (3 x + 7) + \frac{7}{3} ln (3 x + 7) - x - \frac{7}{3}

Add a constant to the solution

= x ln (3 x + 7) + \frac{7}{3} ln (3 x + 7) - x - \frac{7}{3} + C

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What is the integral of ln(3x+7) ?
The integral of ln(3x+7) is xln(3x+7)+7/3 ln(3x+7)-x-7/3+C