The area enclosed by the curves x y + 4 y = 16 and x + y = 6 is equal to:

Given: $xy+4y=16,x+y=6$

$\Rightarrow y\left(x+4\right)=16, x+y=6$

$\Rightarrow \left(6-x\right)\left(x+4\right)=16$

$\Rightarrow 6x+24-x^2-4x-16=0$

$\Rightarrow -x^2+2x+8=0$

$\Rightarrow x^2-2x-8=0$

$\Rightarrow \left(x-4\right)\left(x+2\right)=0$

$\Rightarrow x=4, -2$

So, the required area is given by,

$A=\underset{-2}{\overset{4}{\int }}\left(\left(6-x\right)-\left(\frac{16}{x+4}\right)\right)dx$

$\Rightarrow A={\left[6x-\frac{x^2}{2}-16\log \left|x+4\right|\right]}_{-2}^4$

$\Rightarrow A=\left(24-8-16\log \left|8\right|\right)-\left(-12-2-16\log 2\right)$

$\Rightarrow A=16-16\log \left|8\right|+14+16\log 2$

$\Rightarrow A=30-32\ln 2$