Question

Integrala de la 0 la 1 din x/x+3

Answer 1

$\int\limits^1_0 { \dfrac{x}{x+3} } \, dx = \int\limits^1_0 { \dfrac{x+3-3}{x+3} } \, dx = \int\limits^1_0 { \Big(\dfrac{x+3}{x+3} }- \dfrac{3}{x+3} \Big)\, dx = \\ \\ = \int\limits^1_0 { \Big(1 - \dfrac{3}{x+3} }\Big) \, dx = \int\limits^1_0 { 1} \, dx - \int\limits^1_0 { \dfrac{3}{x+3} } \, dx= x\Big|_0^1 - 3 \int\limits^1_0 { \dfrac{1}{x+3} } \, dx = \\ \\ =1-0 - 3 \cdot \Big(ln(x+3)\Big)\Big|_0^1 = 1 -3\cdot \Big(ln (1+3) - ln(0+3)\Big)= \\ \\ =1-3\cdot (ln4-ln3) = 1 - 3\cdot ln \dfrac{4}{3}$