Question

f(x)=$xln^{2} x$

Sa se calculeze:

$\int\limits^e_1 {f( \frac{1}{x} )} \, dx$

Answer 1

$\int\limits^e_1 { \frac{1}{x}\cdot\ln^2 \frac{1}{x} } \, dx = \\ \int\limits^e_1 {(\ln x)'\ln^2x^{-1}} \, dx= \\ \int\limits^e_1 {(\ln x)'\ln (x^{-1})\ln (x^{-1})} \, dx = \\ \int\limits^e_1 {(\ln x)'(-\ln x)(-\ln x)} \, dx = \\ \int\limits^e_1 {(\ln x)'\ln^2x} \, dx= \\ \ln x\ln^2x- \int\limits^e_1 {(\ln x)(2\ln x)( \frac{1}{x}) } \, dx = \\ \ln^3x- \int\limits^e_1 { \frac{2\ln^2x}{x} } \, dx =\\ \ln^3x-2 \int\limits^e_1 { \frac{1}{x}\ln^2x } \, dx\\ Se\ observa\ ca\ integrala\ se\ repeta.$

$Notam\ integrala\ ceruta\ cu\ I\ si\ am\ obtinut: \\ I=\ln^3x-2I\\3I=\ln^3x\\I= \frac{\ln^3x}{3}$
$Aplicam\ capetele\ de\ integrare: \\ I= \frac{\ln^3x}{3}|_1^e= \frac{1}{3}$