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

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

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integral from 1/e to e of x|ln(x)|

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

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

Decimal

2.24576 \dots

Solution steps

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

Apply u-substitution

: \int_{- 1}^{1} e^{2 u} ∣ u ∣ d u

= \int_{- 1}^{1} e^{2 u} ∣ u ∣ d u

Eliminate Absolutes

= \int_{- 1}^{0} e^{2 u} (- u) d u + \int_{0}^{1} e^{2 u} u d u

\int_{- 1}^{0} e^{2 u} (- u) d u = - \frac{- e ^{2} + 3}{4 e ^{2}}

\int_{0}^{1} e^{2 u} u d u = \frac{e ^{2} + 1}{4}

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

Simplify

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

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

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