Ensino Superior ⇒ Integração Logarítima Tópico resolvido

[carlosa]

Podes me ajudar a encontrar a solução para seguinte integral?
[tex3]\int\limits_{1}^{e}\int\limits_{1}^{e}ln(xy)dydx[/tex3]

[candre]

da pra fazer assim
[tex3]I=\int\limits_1^e\int\limits_1^e\ln xydydx~~~~~~~~\left(\text{integracao por partes com }u=\ln xy,du=\frac{dy}{y},dv=dy,v=y\right)\\
I=\int\limits_1^ey\ln xy\bigg|_{y=1}^{y=e}-\int\limits_1^edydx\\
I=\int\limits_1^ee\ln xe-\ln x-e+1dx\\
I=e\int\limits_1^e\ln xedx-\int\limits_1^e\ln xdx+(1-e)\int\limits_1^edx\\
I=e\left(x\ln xe\bigg|_1^e-\int\limits_1^edx\right)-\left(x\ln x\bigg|_1^e-\int\limits_1^edx\right)+(1-e)(e-1)\\
I=e\left(e\ln e^2-\bcancel{\ln e}-e+\bcancel{1}\right)-\left(\bcancel{e\ln e}-\cancel{\ln 1}-\bcancel{e}+1\right)+(1-e)(e-1)\\
I=e(2e\ln e-e)-1-(e-1)^2\\
I=e^2(2-1)-1-(e-1)^2\\
I=e^2-1-(e-1)^2\\
I=e^2-1-(e^2-2e+1)\\
I=\cancel{e^2}-1-\cancel{e^2}+2e-1\\
I=2e-2=2(e-1)[/tex3]

anotações

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[tex3]x\in\mathbb{R}_+,y\in\mathbb{R}_+[/tex3]
[tex3]u=\ln xy\\
\frac{du}{dx}=\frac{y}{xy}=\frac{1}{x}\\
\frac{du}{dy}=\frac{x}{xy}=\frac{1}{y}[/tex3]
[tex3]\ln x^n=n\ln x[/tex3]
[tex3]\ln xy=\ln x+\ln y[/tex3]

ou por
[tex3]I=\int\limits_1^e\int\limits_1^e\ln xydydx\\
I=\int\limits_1^e\int\limits_1^e\ln x+\ln ydydx\\
I=\int\limits_1^e\int\limits_1^e\ln xdydx+\int\limits_1^e\int\limits_1^e\ln ydydx\\
I=\int\limits_1^e\ln x\int\limits_1^edydx+\int\limits_1^edx\int\limits_1^e\ln ydy\\
I=\int\limits_1^e\ln xdx\int\limits_1^edy+\int\limits_1^edx\int\limits_1^e\ln ydy\\
I=2\int\limits_1^e\ln xdx\int\limits_1^edy\\
I=2\left(x\ln x\bigg|_1^e-\int\limits_1^e\right)(e-1)\\
I=2\left(\cancel{e\ln e}-\cancel{\ln1}-\bcancel{e}+1\right)(e-1)\\
I=2(e-1)[/tex3]