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Problema politehnica

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



Problema Politehnica Salut Aveti Vreo Idee La Problema Numarul 947 Am Incercat Prin Parti Insa Numi Iasa Deloc class=

Răspuns :

[tex] \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) = [/tex]

[tex]= \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 \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_~(+) [/tex]

[tex]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 [/tex]

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