What is the integral of ln(3x+9) ?

The integral of ln(3x+9) is (ln(3(x+3))-1)(x+3)+C

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integral of ln(3x+9)

\int ln (3 x + 9) d x

(ln (3 (x + 3)) - 1) (x + 3) + C

Solution steps

\int ln (3 x + 9) 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 + 9

= \frac{1}{3} ((3 x + 9) ln (3 x + 9) - (3 x + 9))

Simplify

\frac{1}{3} ((3 x + 9) ln (3 x + 9) - (3 x + 9)) : (ln (3 (x + 3)) - 1) (x + 3)

= (ln (3 (x + 3)) - 1) (x + 3)

Add a constant to the solution

= (ln (3 (x + 3)) - 1) (x + 3) + C

\int \frac{sin ( x )}{cos ^{2} ( x ) + cos ( x ) - 2} d x

\frac{s ^{2} - 3}{s + 1 \cdot s + 2}

x \to 0 lim (x^{5} e^{- x^{4}})

\frac{\partial}{\partial x} (e^{- 8 t} cos (π x))

- x + 5, x^{2}

What is the integral of ln(3x+9) ?
The integral of ln(3x+9) is (ln(3(x+3))-1)(x+3)+C