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If the value of the integral

$\displaystyle\int_{0}^{\infty}\dfrac{(x^{2}+4)\ln x}{x^{4}+16}\text{ }\text{d}x$

can be expressed as $\dfrac{\pi\ln a}{b\sqrt{c}}$ where $a$ is a prime number and $c$ is square free find $a+b+c$.

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