Question

integrala de la 1 la e din f(x)/x , f(x)=lnx/x​

Answer 1

$f(x) = \frac{lnx}{x}$

$\int_{1}^{e} \frac{f(x)}{x} dx= ?$

$\int_{1}^{e} \frac{ \frac{lnx}{x} }{x} dx = \int_{1}^{e} \frac{lnx}{x} \times \frac{1}{x} dx = \int_{1}^{e} \frac{lnx}{ {x}^{2} } dx$

$\int \frac{lnx}{ {x}^{2} } dx$

$f(x) = lnx = > f'(x) = (lnx)' = \frac{1}{x}$

$g'(x) = \frac{1}{ {x}^{2} } = {x}^{ - 2} = > g(x) = \int {x}^{ - 2} dx = \frac{ {x}^{ - 2 + 1} }{ - 2 + 1} = \frac{ {x}^{ - 1} }{ - 1} = \frac{ \frac{1}{x} }{ - 1} = - \frac{1}{x}$

$\int \frac{lnx}{ {x}^{2} } dx =f(x) \times g(x)-\int f'(x) \times g(x)dx= lnx \times ( - \frac{1}{x} ) - \int \frac{1}{x} \times ( - \frac{1}{x} )dx$

$= - \frac{lnx}{x} - \int - \frac{1}{ {x}^{2} } dx = - \frac{lnx}{x} + \int \frac{1}{ {x}^{2} }dx$

$= - \frac{lnx}{x} + \int {x}^{ - 2} dx = - \frac{lnx}{x} + \frac{ {x}^{ - 2 + 1} }{ - 2 + 1}$

$= - \frac{lnx}{x} + \frac{ {x}^{ - 1} }{ - 1} = - \frac{lnx}{x} + \frac{ \frac{1}{x} }{ - 1} = - \frac{lnx}{x} - \frac{1}{x} = \frac{ - lnx - 1}{x} + C$

$\int_{1}^{e} \frac{lnx}{ {x}^{2} }dx = \frac{ - lnx - 1}{x} |_{1}^{e}$

$= \frac{ - lne - 1}{e} - \frac{ - ln1 - 1}{1} = \frac{ - 1 - 1}{e} + 0 + 1 = - \frac{2}{e} + 1$