Reportar fallo

## Integral Doble

We integrate the function ${{y}\over{x}}+{{x}\over{y}}$ on the set given by $1 \leq y \leq 3$ and $1 \leq x \leq 2$ . $$I=\iint_{M}f\,\mathrm{d}x\mathrm{d}y\ ,\qquad\qquad f= {{y}\over{x}}+{{x}\over{y}} \ , \qquad\qquad M=\begin{cases} 1 \leq y \leq 3 \\ 1 \leq x \leq 2 \end{cases}$$

$\displaystyle I=\int_{ 1 }^{ 3 }\int_{ 1 }^{ 2 } {{y}\over{x}}+{{x}\over{y}} \;\textrm{d}x\,\textrm{d}y$

$\displaystyle \phantom{I}=\int_{ 1 }^{ 3 }\left[ \ln \left(x\right)\,y+{{x^2}\over{2\,y}} \right]_{ 1 }^{ 2 } \;\textrm{d}y$

$\displaystyle \phantom{I}=\int_{ 1 }^{ 3 } {{1}\over{2}}\,2^2\,y^ {- 1 }+\ln \left(2\right)\,y-\left({{1 }\over{2}}\,1^2\,y^ {- 1 }+\ln \left(1\right)\,y\right) \;\textrm{d}y$ (substituting limits)

$\displaystyle \phantom{I}={\int_{ 1 }^{ 3 } \ln \left(2\right)\,y+{{3}\over{2\,y}} \;\textrm{d}y}$ (simplificación)

$\displaystyle \phantom{I}=\left[ {{3\,\ln \left(y\right)+\ln \left(2\right)\,y^2}\over{2}} \right]_{ 1 }^{ 3 }$

$\displaystyle \phantom{I}= {{3\,\ln \left(3\right)+9\,\ln \left(2\right)}\over{2}}-{{\ln \left(2\right)}\over{2}}$ (substituting limits)

$\displaystyle \phantom{I}= {{3\,\ln \left(3\right)+8\,\ln \left(2\right)}\over{2}}$ (simplificación)

Region for integration