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

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

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integral of xln(x-1)

\int x ln (x - 1) d x

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

Solution steps

\int x ln (x - 1) d x

Apply u-substitution

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

= \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

= 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

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{x ^{2}}{5 + 8 x}

f (x) = (x^{5} - x^{3} + 3)

\int \frac{1}{x ( 5 + 3 ln ( x ) )} d x

\int 3^{- 2 x} d x

\int \frac{2 x ^{2} - 2 x - 8}{x ^{2} ( x - 4 )} d x

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