# Integral int_(1/e)^e ln(x)^2018/(x+1) dx ?

Apr 3, 2018

$F \left(e\right) - F \left(\frac{1}{e}\right) = \frac{2}{2019}$

#### Explanation:

$F \left(x\right) = {\int}_{\frac{1}{e}}^{e} {\ln}^{2018} \frac{x}{x + 1} \mathrm{dx} = {\ln}^{2019} \frac{x}{2019}$

Evaluate $F \left(e\right) - F \left(\frac{1}{e}\right)$

$F \left(e\right) = {\ln}^{2019} \frac{e}{2019} = {1}^{2019} / 2019 = \frac{1}{2019}$

$F \left(\frac{1}{e}\right) = {\ln}^{2019} \frac{\frac{1}{e}}{2019} = {\left(- 1\right)}^{2019} / 2019 = - \frac{1}{2019}$

Finally: $F \left(e\right) - F \left(\frac{1}{e}\right) = \frac{1}{2019} - \left(- \frac{1}{2019}\right) = \frac{2}{2019}$

Apr 3, 2018

${\int}_{\frac{1}{e}}^{e} {\left(L n x\right)}^{2018} / \left(x + 1\right) \cdot \mathrm{dx} = \frac{1}{2019}$

#### Explanation:

$I = {\int}_{\frac{1}{e}}^{e} {\left(L n x\right)}^{2018} / \left(x + 1\right) \cdot \mathrm{dx}$

After using $x = \frac{1}{y}$ and $\mathrm{dx} = - \frac{\mathrm{dy}}{y} ^ 2$ transform,

$I = {\int}_{e}^{\frac{1}{e}} {\left(L n \left(\frac{1}{y}\right)\right)}^{2018} / \left(\frac{1}{y} + 1\right) \cdot \left(- \frac{\mathrm{dy}}{y} ^ 2\right)$

=${\int}_{e}^{\frac{1}{e}} {\left(- L n y\right)}^{2018} / \left(\frac{y + 1}{y}\right) \cdot \left(- \frac{\mathrm{dy}}{y} ^ 2\right)$

=${\int}_{e}^{\frac{1}{e}} {\left(L n y\right)}^{2018} / \left({y}^{2} + y\right) \cdot \left(- \mathrm{dy}\right)$

=${\int}_{\frac{1}{e}}^{e} {\left(L n y\right)}^{2018} / \left({y}^{2} + y\right) \cdot \mathrm{dy}$

=${\int}_{\frac{1}{e}}^{e} {\left(L n x\right)}^{2018} / \left({x}^{2} + x\right) \cdot \mathrm{dx}$

After summing 2 integrals,

$2 I = {\int}_{\frac{1}{e}}^{e} {\left(L n x\right)}^{2018} / \left(x + 1\right) \cdot \mathrm{dx}$+${\int}_{\frac{1}{e}}^{e} {\left(L n x\right)}^{2018} / \left({x}^{2} + x\right) \cdot \mathrm{dx}$

=${\int}_{\frac{1}{e}}^{e} \frac{x \cdot {\left(L n x\right)}^{2018}}{{x}^{2} + x} \cdot \mathrm{dx}$+${\int}_{\frac{1}{e}}^{e} {\left(L n x\right)}^{2018} / \left({x}^{2} + x\right) \cdot \mathrm{dx}$

=${\int}_{\frac{1}{e}}^{e} \frac{\left(x + 1\right) \cdot {\left(L n x\right)}^{2018} \cdot \mathrm{dx}}{x \cdot \left(x + 1\right)}$

=${\int}_{\frac{1}{e}}^{e} \frac{{\left(L n x\right)}^{2018} \cdot \mathrm{dx}}{x}$

=${\left[{\left(L n x\right)}^{2019} / 2019\right]}_{\frac{1}{e}}^{e}$

=$\frac{2}{2019}$

Thus,

$I = {\int}_{\frac{1}{e}}^{e} {\left(L n x\right)}^{2018} / \left(x + 1\right) \cdot \mathrm{dx} = \frac{1}{2019}$