A calculus problem by Aly Ahmed

The above integrand can be written as $1 + \frac{21}{8} (\frac{1}{x-4} - \frac{1}{x+4})$ , which integrates into:

$\int_{0}^{1} 1 + \frac{21}{8} (\frac{1}{x-4} - \frac{1}{x+4}) dx = x + \frac{21}{8}[\ln(x-4) - \ln(x+4)]|_{0}^{1} = \boxed{1 + \frac{21}{8} \cdot ln( \frac{3}{5} )}.$

$\begin{aligned} I & = \int_0^1 \frac {x^2+5}{x^2-16} dx \\ & = \int_0^1 \frac {x^2-16+21}{x^2-16} dx \\ & = \int_0^1 \left(1 + \frac {21}{(x-4)(x+4)} \right) dx \\ & = \int_0^1 \left(1 + \frac {21}8 \left(\frac 1{x-4} - \frac 1{x+4} \right) \right) dx \\ & = x + \frac {21}8 \ln \left|\frac {x-4}{x+4} \right|\ \bigg|_0^1 \\ & = 1 + \frac {21}8 \left(\ln \frac 35 - \ln \frac 44 \right) \\ & = 1 + \frac {21(\ln 3-\ln 5)}8 \approx \boxed{-0.341} \end{aligned}$