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

Given: $xy + 4y = 16, x + y = 6$

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

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

$⇒ 6x + 24 - {x}^{2} - 4x - 16 = 0$

$⇒ - {x}^{2} + 2x + 8 = 0$

$⇒ {x}^{2} - 2x - 8 = 0$

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

$⇒ x = 4, - 2$

So, the required area is given by,

$A = \mathop{∫}_{- 2}^{4}\left(\left(6 - x\right) - \left(\dfrac{16}{x + 4}\right)\right)dx$

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

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

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

$⇒ A = 30 - 32\ln 2$