Reportar fallo

## Integral Doble

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

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

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

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

$\displaystyle \phantom{I}={\int_{ -2 }^{ 2 } {{7}\over{2\,x}}+2 \;\textrm{d}x}$ (simplificación)

$\displaystyle \phantom{I}=\left[ {{7\,\ln \left(\left| x\right| \right)+4\,x+7\,i\,{\rm atan2} \left(0 , x\right)}\over{2}} \right]_{ -2 }^{ 2 }$

$\displaystyle \phantom{I}= {{7\,\ln \left(2\right)+8}\over{2}}-{{7\,\ln \left(2\right)+7\,i \,\pi-8}\over{2}}$ (substituting limits)

$\displaystyle \phantom{I}= -{{7\,i\,\pi-16}\over{2}}$ (simplificación)

Region for integration