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

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

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

\int_{1}^{e} x \cdot 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}

2 sin (x), 4 cos (x)

\frac{d}{d x} (x tan (x))

\int \frac{1}{9 + 4 x ^{2}} d x

\int \frac{2}{5} - \frac{9}{x} d x

x d y - (y + x^{3} e^{x^{2}}) d x = 0

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