What is the integral of tln(t-1) ?

The integral of tln(t-1) is 1/2 ln(t-1)(t-1)^2+tln(t-1)-ln(t-1)-1/4 (t-1)^2-t+1+C

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integral of tln(t-1)

\int t ln (t - 1) d t

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

Solution steps

\int t ln (t - 1) d t

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 = t - 1

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

Simplify

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

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

Add a constant to the solution

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

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