Question

Calculati:
$\int\limits^1_2 {ln(1+\frac{2}{x})}\, dx$

Va rog sa explicati ce metode ati folosit si sa scrieti mai detaliat. Multumesc!

Answer 1

∫₁² ln(1 + 2/x) dx =

= ∫₁² 1•ln(1 + 2/x) dx

= ∫₁² x'•ln(1 + 2/x) dx

(Folosesc integrarea prin părți:

∫ f'•g dx = f•g - ∫ f•g' dx)

= x•ln(1 + 2/x) |₁² - ∫₁² x•[ln(1 + 2/x)]' dx

= 2ln2 - ln3 - ∫₁² x•(-2/x²)/(1 + 2/x) dx

= ln4 - ln3 + ∫₁² 2/(x+2) dx

= ln(4/3) + 2∫₁² 1/(x+2) dx

= ln(4/3) + 2∫₁² (x+2)'/(x+2) dx

= ln(4/3) + 2ln(x+2)|₁²

= ln(4/3) + 2ln(4) - 2ln(3)

= ln(4/3) + 2ln(4/3)

= 3ln(4/3)