What is the integral from 1 to e of zln(z) ?

The integral from 1 to e of zln(z) is (e^2+1)/4

integral from 1 to e of zln(z)

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

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

Decimal

2.09726 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

= [\frac{1}{2} (z^{2} ln (z) - \frac{z ^{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}

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