# How to solve this? int_1^3 (x^2-16) /(x-4

Mar 17, 2018

${\int}_{1}^{3} \frac{{x}^{2} - 16}{x - 4} \mathrm{dx} = 12$

#### Explanation:

${\int}_{1}^{3} \frac{{x}^{2} - 16}{x - 4} \mathrm{dx}$

= ${\int}_{1}^{3} \frac{\left(x + 4\right) \left(x - 4\right)}{x - 4} \mathrm{dx}$

= ${\int}_{1}^{3} \left(x + 4\right) \mathrm{dx}$

= ${\left[{x}^{2} / 2 + 4 x\right]}_{1}^{3}$

= $\left[{3}^{2} / 2 + 4 \cdot 3 - {1}^{2} / 2 - 4 \cdot 1\right]$

= $\frac{9}{2} + 12 - \frac{1}{2} - 4$

= $12$