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

The integral of 2xln(x-2) is 2(ln(x-2)(1/2 (x-2)^2+2(x-2))-1/4 (x^2+4x-12))+C

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integral of 2xln(x-2)

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

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

Solution steps

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

Apply u-substitution

: \int (u + 2) ln (u) d u

= 2 \cdot \int (u + 2) ln (u) d u

Apply Integration By Parts

: ln (u) (\frac{u ^{2}}{2} + 2 u) - \int \frac{1}{2} (u + 4) d u

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

\int \frac{1}{2} (u + 4) d u = \frac{1}{2} (\frac{u ^{2}}{2} + 4 u)

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

Substitute back

u = x - 2

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{6} \frac{3}{x ^{2} + 2 x + 1} d x

\frac{\partial}{\partial x} (e^{x + 9 y})

y = cos (x), y = x + 1, 0, \frac{π}{2}

x \to 0 lim (+ (x ln (x)))

∣ 2 x ∣, 2 x^{2} + x - 4

What is the integral of 2xln(x-2) ?
The integral of 2xln(x-2) is 2(ln(x-2)(1/2 (x-2)^2+2(x-2))-1/4 (x^2+4x-12))+C