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

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

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integral from 1 to e of xln(x)

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

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

Decimal Notation

2.09726 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

2 \cdot \frac{x ^{2}}{4} = \frac{x ^{2}}{2}

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

Compute the boundaries:

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

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

Simplify

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

\frac{d ^{2} y}{d x ^{2}} - y = 0

x \to 0 lim (3 (x)^{\frac{x}{2}})

\frac{( 2 s + 5 )}{s ^{2} + 6 s + 34}

\int x e^{x + 2} d x

n = 1 \sum \infty \frac{( x - 1 ) ^{n}}{n}

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