Bonjour
Linoush99
Dérivée
[tex][10\times(\dfrac{1+\ln x}{x})]'=10\times[\dfrac{1+\ln x}{x}]'\\\\\\=10\times[\dfrac{(1+\ln x)'\times x-(1+\ln x)\times x'}{x^2}]\\\\\\=10\times[\dfrac{(0+\dfrac{1}{x})\times x-(1+\ln x)\times1}{x^2}]\\\\\\=10\times[\dfrac{\dfrac{1}{x}\times x-(1+\ln x)}{x^2}]\\\\\\=10\times[\dfrac{1-1-\ln x}{x^2}]\\\\\\=10\times[\dfrac{-\ln x}{x^2}]\\\\\\=\dfrac{-10\ln x}{x^2}\\\\\\\Longrightarrow\boxed{[10\times(\dfrac{1+\ln x}{x})]'=\dfrac{-10\ln x}{x^2}}[/tex]
Primitive
[tex]\int10\times(\dfrac{1+\ln x}{x})\,dx=10\times\int\dfrac{1+\ln x}{x}\,dx\\\\\\=10\times\int(\dfrac{1}{x}+\dfrac{\ln x}{x})\,dx\\\\\\=10\times\int\dfrac{1}{x}\,dx+10\times\int\dfrac{\ln x}{x}\,dx\\\\\\=10\times\int\dfrac{1}{x}\,dx+10\times\int(\dfrac{1}{x}\times\ln x)\,dx\\\\\\=10\times\ln x+10\times\dfrac{\ln^2x}{2}+constante\\\\\\=10\times\ln x+5\times\ln^2x+constante\\\\\\=\boxed{5\ln x(2+\ln x)+constante}[/tex]
Par conséquent,
une primitive de [tex]10\times(\dfrac{1+\ln x}{x})[/tex] est [tex]\boxed{5\ln x(\ln x+2)}[/tex]
