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

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

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integral from t to 2 of-xln(x-1)

\int_{t}^{2} - x ln (x - 1) d x

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

Solution steps

\int_{t}^{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

= - \int_{t}^{2} x ln (x - 1) d x

Apply u-substitution

: \int_{t - 1}^{1} (u + 1) ln (u) d u

= - \int_{t - 1}^{1} (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]_{t - 1}^{1}

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

\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)]_{t - 1}^{1}

Simplify

- [ln (u) (\frac{u ^{2}}{2} + u) - (\frac{u ^{2}}{4} + u)]_{t - 1}^{1} : - [\frac{1}{2} u^{2} ln (u) + u ln (u) - \frac{u ^{2}}{4} - u]_{t - 1}^{1}

= - [\frac{1}{2} u^{2} ln (u) + u ln (u) - \frac{u ^{2}}{4} - u]_{t - 1}^{1}

Compute the boundaries:

- \frac{5}{4} - \frac{1}{2} t^{2} ln (t - 1) + \frac{1}{2} ln (t - 1) + t + \frac{( t - 1 ) ^{2}}{4} - 1

= - (- \frac{5}{4} - \frac{1}{2} t^{2} ln (t - 1) + \frac{1}{2} ln (t - 1) + t + \frac{( t - 1 ) ^{2}}{4} - 1)

Simplify

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

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