A calculus problem by Aly Ahmed

There is a direct method, where we can avoid integration.

$\begin{aligned} f(x) & = \int_{\ln x}^{\ln(x^2+3)} \frac {dt}{3+e^t} & \small \blue{\text{For }f(x) = \int_k^{h(x)}g(t)\ dt \text{, where }k \text{ is a constant,}} \\ \implies f'(x) & = \frac 1{3+e^{\ln(x^2+3)}} \cdot \frac {d \ln(x^2+3)}{dx} - \frac 1{3+e^{\ln x}} \cdot \frac {d \ln x}{dx} & \small \blue{\implies f'(x) = g(h(x)) h'(x)} \\ & = \frac 1{x^2+6} \cdot \frac {2x}{x^2+3} - \frac 1{x+3} \cdot \frac 1x \\ \implies f'(2) & = \frac 4{70} - \frac 1{10} = - \frac 3{70} \end{aligned}$

Therefore, $a=\boxed{70}$ .

By integrating we get $f(x)=\dfrac {1}{3}\ln \dfrac{(x+3)(x^2+3)}{x(x^2+6)}$

So $f'(x) =\dfrac {-3x^4+6x^3-9x^2-54}{3x(x+3)(x^2+3)(x^2+6)}$

$\implies f'(2)=\dfrac{48-48-36-54}{3\times 2\times 5\times 7\times 10}=-\dfrac{3}{70}$

Hence $a=\boxed {70}$ .