What is the integral of ln(4-x) ?

The integral of ln(4-x) is -4ln(4-x)+xln(4-x)+4-x+C

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integral of ln(4-x)

\int ln (4 - x) d x

- 4 ln (4 - x) + x ln (4 - x) + 4 - x + C

Solution steps

\int ln (4 - 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 = 4 - x

= - ((4 - x) ln (4 - x) - (4 - x))

Simplify

- ((4 - x) ln (4 - x) - (4 - x)) : - 4 ln (4 - x) + x ln (4 - x) + 4 - x

= - 4 ln (4 - x) + x ln (4 - x) + 4 - x

Add a constant to the solution

= - 4 ln (4 - x) + x ln (4 - x) + 4 - x + C

\frac{d f ^{- 1}}{d x} f (x) = 3 x^{3} - 5 x^{2} - 2, x = 1272

\frac{1 + x}{1 + e ^{x}}

x = 8 y^{2}, x = 20 + 3 y^{2}

\frac{d}{d x} ((x^{2} + 5 x + 2)^{7})

\int x^{3} 64 - x^{2} d x

What is the integral of ln(4-x) ?
The integral of ln(4-x) is -4ln(4-x)+xln(4-x)+4-x+C