What is the integral from 1 to x of u*ln(u) ?

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

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integral from 1 to x of u*ln(u)

\int_{1}^{x} u \cdot ln (u) d u

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

Solution steps

\int_{1}^{x} u ln (u) d u

Apply Integration By Parts

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

\int \frac{u}{2} d u = \frac{u ^{2}}{4}

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

Compute the boundaries:

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

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

Simplify

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

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