What is the integral of ln(2-x) ?

The integral of ln(2-x) is -2ln(2-x)+xln(2-x)+2-x+C

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integral of ln(2-x)

\int ln (2 - x) d x

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

Solution steps

\int ln (2 - 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 = 2 - x

= - ((2 - x) ln (2 - x) - (2 - x))

Simplify

- ((2 - x) ln (2 - x) - (2 - x)) : - 2 ln (2 - x) + x ln (2 - x) + 2 - x

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

Add a constant to the solution

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

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

\frac{d ^{2}}{d x ^{2}} (x^{\frac{5}{2}} e^{x})

\int cos (0)

\frac{d}{d x} ((3 x + 5)^{3})

\frac{\partial}{\partial x} (x^{4} y - 4 x^{5} y^{2})

What is the integral of ln(2-x) ?
The integral of ln(2-x) is -2ln(2-x)+xln(2-x)+2-x+C