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

The integral from 1 to e of (x+1)ln(x) is (e^2+5)/4

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

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

\frac{e ^{2} + 5}{4}

Decimal

3.09726 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{1}{2} (x + 2) d x = \frac{x ^{2}}{4} + x

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

Simplify

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

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

Compute the boundaries:

\frac{e ^{2}}{4} + \frac{5}{4}

= \frac{e ^{2}}{4} + \frac{5}{4}

Simplify

= \frac{e ^{2} + 5}{4}

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What is the integral from 1 to e of (x+1)ln(x) ?
The integral from 1 to e of (x+1)ln(x) is (e^2+5)/4