Question

Problema politehnica

Salut, aveti vreo idee la problema numarul 947 ? Am incercat prin parti insa nu-mi iasa deloc...

Answer 1

$\int\limits^3_1 \dfrac{ \ln x}{\ln \Big(x(4-x)\Big)} \, dx \\ \\ Facem schimbarea de variabila:\quad t = 4-x \Rightarrow dt = -dx \Rightarrow dx = -dt \\ x = 1 \Rightarrow t = 4-1 \Rightarrow t=3 \\ x = 3 \Rightarrow t = 4-3 \Rightarrow t = 1 \\ t = 4-x \Rightarrow x = 4-t\\ \\ Integrala devine: \\ \\ \int\limits^1_3 \dfrac{ \ln (4-t)}{\ln \Big(\big(4-t\big)\big(4-(4-t)\big)\Big)} \, \cdot(-dt) =$

$= \int\limits^1_3 \dfrac{ \ln (4-t)}{\ln \Big(\big(4-t\big)\big(4-4+t\big)\Big)} \, \cdot(-dt) = \int\limits^1_3 \dfrac{ \ln (4-t)}{\ln \Big(t\cdot(4-t)\Big)} \, \cdot(-dt) \\ \\ =\int\limits^3_1 \dfrac{ \ln (4-t)}{\ln \Big(t\cdot(4-t)\Big)} \, dt = \int\limits^3_1 \dfrac{ \ln (4-x)}{\ln \Big(x\cdot(4-x)\Big)} \, dx \\ \\\\ I = \int\limits^3_1 \dfrac{ \ln x}{\ln \Big(x(4-x)\Big)} \, dx \\ \\ I =\int\limits^3_1 \dfrac{ \ln (4-x)}{\ln \Big(x(4-x)\Big)} \, dx \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_~(+)$

$2I = \int\limits^3_1 \dfrac{ \ln x}{\ln \Big(x(4-x)\Big)} \, dx +\int\limits^3_1 \dfrac{ \ln (4-x)}{\ln \Big(x(4-x)\Big)} \, dx \\ \\ 2I = \int\limits^3_1 \left(\dfrac{ \ln x}{\ln \Big(x(4-x)\Big)} + \dfrac{ \ln (4-x)}{\ln \Big(x(4-x)\Big)}\right) \, dx \\ \\ 2I = \int\limits^3_1\left \dfrac{ \ln x+\ln(4-x)}{\ln \Big(x(4-x)\Big)}\, dx \\ \\ 2I = \int\limits^3_1\left \dfrac{ \ln\Big(x(4-x)\Big)}{\ln \Big(x(4-x)\Big)}\, dx \\ \\ 2I = \int\limits^3_1 1\, dx\\ \\ 2I = x\Big|_1^3 \\ \\ 2I = 3-1$

$2I = 2 \\ \\ I=1 \\ \\ \Rightarrow \boxed{\int\limits^3_1 \dfrac{ \ln x}{\ln \Big(x(4-x)\Big)} \, dx = 1}$