What is the integral of 2xln(x-1) ?

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

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integral of 2xln(x-1)

\int 2 x ln (x - 1) d x

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

Solution steps

\int 2 x ln (x - 1) 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 - 1) d x

Apply u-substitution

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

= 2 \cdot \int (u + 1) ln (u) d u

Apply Integration By Parts

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

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

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

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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