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

解答

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

解答

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

+1

十进制

3.51000 \dots

求解步骤

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

乘以

x + 1

的共轭根式

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

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

找到无定义的点（奇点）:

x = 1

如果存在

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

无定义

, \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

化简

- \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

作图

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