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

The integral from 1/e to e of |ln(x)|(x+1) is (e^4+2e^2-8e-3)/(4e^2)+2

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

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

\frac{e ^{4} + 2 e ^{2} - 8 e - 3}{4 e ^{2}} + 2

Decimal Notation

3.51000 \dots

Solution steps

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

Multiply by the conjugate of

x + 1 : \frac{x ^{2} - 1}{x - 1}

= \int_{\frac{1}{e}}^{e} ∣ ln (x) ∣ \frac{x ^{2} - 1}{x - 1} d x

Find undefined (singularity) points:

x = 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{\frac{1}{e}}^{1} ∣ ln (x) ∣ \frac{x ^{2} - 1}{x - 1} d x + \int_{1}^{e} ∣ ln (x) ∣ \frac{x ^{2} - 1}{x - 1} d x

\int_{\frac{1}{e}}^{1} ∣ ln (x) ∣ \frac{x ^{2} - 1}{x - 1} d x = - \frac{2}{e} - \frac{- e ^{2} + 3}{4 e ^{2}} + 1

\int_{1}^{e} ∣ ln (x) ∣ \frac{x ^{2} - 1}{x - 1} d x = \frac{e ^{2} + 1}{4} + 1

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

Simplify

- \frac{2}{e} - \frac{- e ^{2} + 3}{4 e ^{2}} + 1 + \frac{e ^{2} + 1}{4} + 1 : \frac{e ^{4} + 2 e ^{2} - 8 e - 3}{4 e ^{2}} + 2

= \frac{e ^{4} + 2 e ^{2} - 8 e - 3}{4 e ^{2}} + 2

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What is the integral from 1/e to e of |ln(x)|(x+1) ?
The integral from 1/e to e of |ln(x)|(x+1) is (e^4+2e^2-8e-3)/(4e^2)+2