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

The integral of 2(x-2)*ln(x-2) is x^2ln(x-2)-1/2 x^2+2x-4xln(x-2)+4ln(x-2)-2+C

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integral of 2(x-2)*ln(x-2)

\int 2 (x - 2) \cdot ln (x - 2) d x

x^{2} ln (x - 2) - \frac{1}{2} x^{2} + 2 x - 4 x ln (x - 2) + 4 ln (x - 2) - 2 + C

Solution steps

\int 2 (x - 2) ln (x - 2) d x

Take the constant out:

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

= 2 \cdot \int (x - 2) ln (x - 2) d x

Expand

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

= 2 \cdot \int x ln (x - 2) - 2 ln (x - 2) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int x ln (x - 2) d x - \int 2 ln (x - 2) d x)

\int x ln (x - 2) d x = ln (x - 2) (\frac{1}{2} (x - 2)^{2} + 2 (x - 2)) - \frac{1}{4} (x^{2} + 4 x - 12)

\int 2 ln (x - 2) d x = 2 (x ln (x - 2) - 2 ln (x - 2) - x + 2)

= 2 (ln (x - 2) (\frac{1}{2} (x - 2)^{2} + 2 (x - 2)) - \frac{1}{4} (x^{2} + 4 x - 12) - 2 (x ln (x - 2) - 2 ln (x - 2) - x + 2))

Simplify

2 (ln (x - 2) (\frac{1}{2} (x - 2)^{2} + 2 (x - 2)) - \frac{1}{4} (x^{2} + 4 x - 12) - 2 (x ln (x - 2) - 2 ln (x - 2) - x + 2)) : x^{2} ln (x - 2) - \frac{1}{2} x^{2} + 2 x - 4 x ln (x - 2) + 4 ln (x - 2) - 2

= x^{2} ln (x - 2) - \frac{1}{2} x^{2} + 2 x - 4 x ln (x - 2) + 4 ln (x - 2) - 2

Add a constant to the solution

= x^{2} ln (x - 2) - \frac{1}{2} x^{2} + 2 x - 4 x ln (x - 2) + 4 ln (x - 2) - 2 + C

\int \frac{1}{4 - 6 x} d x

d x + d y + x d y = y d x

f (x) = \frac{2 π - x cos ( x )}{x - 2 π}

x \to + 1 lim (\frac{x - 1}{x ^{2} - x})

x \frac{d y}{d x} = 8 y

What is the integral of 2(x-2)*ln(x-2) ?
The integral of 2(x-2)*ln(x-2) is x^2ln(x-2)-1/2 x^2+2x-4xln(x-2)+4ln(x-2)-2+C