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

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

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integral of ln(3-x)

\int ln (3 - x) d x

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

Solution steps

\int ln (3 - x) d x

Apply u-substitution

: \int - ln (u) d u

= \int - ln (u) d u

Take the constant out:

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

= - \int ln (u) d u

Apply Integration By Parts

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

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

\int 1 d u = u

= - (u ln (u) - u)

Substitute back

u = 3 - x

= - ((3 - x) ln (3 - x) - (3 - x))

Simplify

- ((3 - x) ln (3 - x) - (3 - x)) : - 3 ln (3 - x) + x ln (3 - x) + 3 - x

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

Add a constant to the solution

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

\int_{1}^{2} \frac{x - 1}{2 x} d x

\frac{\partial}{\partial y} (sin (\frac{5 x}{y}))

y = \frac{4}{x}, y = x, y = 4

\int \frac{x}{3 x - 4} d x

\int 5 x^{2} (x^{3} + 1)^{10} d x

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