Answer
$$f\left( t \right) = - \frac{8}{{{{\left( {t + 3} \right)}^2}}}{\text{ and }}a = 1$$
Work Step by Step
$$\eqalign{
& 2 + \int_a^x {f\left( t \right)} dt = \frac{8}{{x + 3}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ 2 \right] + \frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)} dt} \right] = \frac{d}{{dx}}\left[ {\frac{8}{{x + 3}}} \right] \cr
& 0 + f\left( x \right) = - \frac{8}{{{{\left( {x + 3} \right)}^2}}} \cr
& {\text{Let }}x = t \cr
& {\text{ }}f\left( t \right) = - \frac{8}{{{{\left( {t + 3} \right)}^2}}} \cr
& {\text{Substitute }}f\left( t \right) = - \frac{8}{{{{\left( {t + 3} \right)}^2}}}{\text{ into }}\left( {\bf{1}} \right) \cr
& 2 + \int_a^x {\left[ { - \frac{8}{{{{\left( {t + 3} \right)}^2}}}} \right]} dt = \frac{8}{{x + 3}} \cr
& {\text{Integrating}} \cr
& 2 + \left[ {\frac{8}{{t + 3}}} \right]_a^x = \frac{8}{{x + 3}} \cr
& 2 + \frac{8}{{x + 3}} - \frac{8}{{a + 3}} = \frac{8}{{x + 3}} \cr
& {\text{Solve for }}a \cr
& 2 - \frac{8}{{a + 3}} = 0 \cr
& \frac{8}{{a + 3}} = 2 \cr
& 2a + 6 = 8 \cr
& a = 1 \cr
& {\text{Therefore,}} \cr
& f\left( t \right) = - \frac{8}{{{{\left( {t + 3} \right)}^2}}}{\text{ and }}a = 1 \cr} $$
