Problem #2: Complication by simplification

Note that

$\frac{x^2-1}{x^4+3x^2+1}=\frac{1-\frac{1}{x^2}}{x^2+3+\frac{1}{x^2}}=\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2+1},$

and hence by substitution $x+\dfrac{1}{x}=t,$ the integral becomes

$\int_{\frac{13}{6}}^{\frac{26}{5}}\frac{1}{1+t^2}dt,$

and with another substitution $t=\tan p$ where $\tan \alpha=\dfrac{13}{6},~\tan \beta = \dfrac{26}{5},$ the integral becomes

$\int_{\alpha}^{\beta} \frac{\sec^2 p}{1+\tan^2 p}dp =\int_{\alpha}^{\beta} dp = \beta-\alpha,$

so that

$\tan k = \tan (\beta-\alpha) = \frac{\tan \beta - \tan \alpha}{1+\tan \alpha \cdot \tan \beta} = \frac{91}{368}.$

Hence $p+q=91+368=\boxed{459}.$

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Concise version of the solution:

Let $x+\dfrac{1}{x}=\tan p.$ The integral becomes

$\int_{\arctan \frac{13}{6}}^{\arctan \frac{26}{5}} \frac{\sec^2 p}{1+\tan^2 p} dp \\ = \arctan \frac{26}{5}-\arctan \frac{13}{6} \\ = \frac{91}{368}.$

The answer is $\boxed{459}.$