Răspuns:
e-2
Explicație pas cu pas:
[tex]\int\limits^e_1 {ln^2x} \, dx =Punem~x'~in~fata~logaritmului=\int\limits^e_1 {x'*ln^2x} \, dx=Integram~prin~parti=x*ln^2x|^e_1-\int\limits^e_1 {x(ln^2x)'} \, dx=Facem ~calculele~care~se~impun =e*ln^2e-1*ln^21-\int\limits^e_1 {x*2*lnx*\frac{1}{x} } \, dx=[/tex]
[tex]Continuam~calculele~din~fata~integralei~si~simplificam~x~cu~\frac{1}{x} ~in~integrala=e*1^2-1*0^2-\int\limits^e_1 {2lnx} \, dx=Continuam~calculul~si~scoatem din~integrala~2=e-2\int\limits^e_1 {lnx} \, dx =Adaugam~x'~in~fata~logaritmului=e-2\int\limits^e_1 {x'*lnx} \, dx=[/tex]
[tex]Integram~prin~parti=e-2*(x*lnx|^e_1-\int\limits^e_1 {x*(lnx)'} \, dx)=e-2*(e*lne-1*ln1-\int\limits^e_1 {x*\frac{1}{x} } \, dx )=e-2*(e-\int\limits^e_1 {} \, dx )=e-2(e-x|^e_1)=e-2(e-e+1)=e-2[/tex]
