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

The integral from 1 to 2 of (x-1)ln(x) is 1/4

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integral from 1 to 2 of (x-1)ln(x)

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

\frac{1}{4}

Decimal

0.25

Solution steps

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

Apply Integration By Parts

: [ln (x) (\frac{x ^{2}}{2} - x) - \int \frac{1}{2} (x - 2) d x]_{1}^{2}

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

\int \frac{1}{2} (x - 2) d x = \frac{x ^{2}}{4} - x

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

Simplify

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

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

Compute the boundaries:

\frac{1}{4}

= \frac{1}{4}

\frac{d}{d x} (\frac{- 2}{x - 5} - 3)

\frac{d}{d x} (\frac{1}{12 x ^{2}})

\frac{d}{d x} ((1 - x) (x^{2} - 3)^{2})

\int \frac{cos ( t )}{sin ( t )} d t

x^{2} \frac{d y}{d x} + x (x + 2) = e^{x}

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