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

The integral from 1 to e of xln(x)-x is (-e^2+3)/4

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integral from 1 to e of xln(x)-x

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

\frac{- e ^{2} + 3}{4}

Decimal Notation

- 1.09726 \dots

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{1}^{e} x ln (x) d x = \frac{e ^{2} + 1}{4}

\int_{1}^{e} x d x = \frac{e ^{2} - 1}{2}

= \frac{e ^{2} + 1}{4} - \frac{e ^{2} - 1}{2}

Simplify

\frac{e ^{2} + 1}{4} - \frac{e ^{2} - 1}{2} : \frac{- e ^{2} + 3}{4}

= \frac{- e ^{2} + 3}{4}

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What is the integral from 1 to e of xln(x)-x ?
The integral from 1 to e of xln(x)-x is (-e^2+3)/4