What is the integral from 1 to 3 of (x-1)ln(x) ?

The integral from 1 to 3 of (x-1)ln(x) is 3/2 ln(3)

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integral from 1 to 3 of (x-1)ln(x)

\int_{1}^{3} (x - 1) ln (x) d x

\frac{3}{2} ln (3)

Decimal Notation

1.64791 \dots

Solution steps

\int_{1}^{3} (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}^{3}

= [ln (x) (\frac{x ^{2}}{2} - x) - \int \frac{1}{2} (x - 2) d x]_{1}^{3}

\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}^{3}

Simplify

[ln (x) (\frac{x ^{2}}{2} - x) - (\frac{x ^{2}}{4} - x)]_{1}^{3} : [\frac{1}{2} x^{2} ln (x) - x ln (x) - \frac{x ^{2}}{4} + x]_{1}^{3}

= [\frac{1}{2} x^{2} ln (x) - x ln (x) - \frac{x ^{2}}{4} + x]_{1}^{3}

Compute the boundaries:

\frac{3}{2} ln (3)

= \frac{3}{2} ln (3)

\frac{d}{d x} (\frac{4}{3} p (x) i r (x) (x)^{3})

\frac{\partial}{\partial x} (y ln (x^{4} + y^{3} + 2) - 7)

(\frac{4}{3}) p i x^{3}

\frac{d}{d x} ((x^{3} + 2 x)^{4})

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

What is the integral from 1 to 3 of (x-1)ln(x) ?
The integral from 1 to 3 of (x-1)ln(x) is 3/2 ln(3)