What is the integral of ln(6-3x) ?

The integral of ln(6-3x) is -(ln(3(-x+2))-1)(-x+2)+C

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integral of ln(6-3x)

\int ln (6 - 3 x) d x

- (ln (3 (- x + 2)) - 1) (- x + 2) + C

Solution steps

\int ln (6 - 3 x) d x

Apply u-substitution

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

= \int - \frac{1}{3} ln (u) 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 = 6 - 3 x

= - \frac{1}{3} ((6 - 3 x) ln (6 - 3 x) - (6 - 3 x))

Simplify

- \frac{1}{3} ((6 - 3 x) ln (6 - 3 x) - (6 - 3 x)) : - (ln (3 (- x + 2)) - 1) (- x + 2)

= - (ln (3 (- x + 2)) - 1) (- x + 2)

Add a constant to the solution

= - (ln (3 (- x + 2)) - 1) (- x + 2) + C

2 y^{^{''}} - 5 y^{^{'}} + 6 y = 0

cos (\frac{π}{2} x)

\frac{e ^{- s} + s}{s ^{2} + s}

\frac{d}{d x} (\frac{2 ( 3 x ^{2} + 1 )}{( 1 - x ^{2} ) ^{3}})

\int_{- a}^{a} \int_{0}^{1} sin (y) e^{x^{2}} d x d y

What is the integral of ln(6-3x) ?
The integral of ln(6-3x) is -(ln(3(-x+2))-1)(-x+2)+C