$$I = \int {{{{e^x}({x^2} + 1)} \over {{{(x + 1)}^2}}}dx = f(x){e^x} + c} $$

$$ = \int {{{{e^x}({x^2} - 1 + 1 + 1)} \over {{{(x + 1)}^2}}}dx} $$

$$ = \int {{e^x}\left[ {{{x - 1} \over {x + 1}} + {2 \over {{{(x + 1)}^2}}}} \right]dx} $$

$$ = {e^x}\left( {{{x - 1} \over {x + 1}}} \right) + c$$

$$\therefore$$ $$f(x) = {{x - 1} \over {x + 1}}$$

$$f(x) = 1 - {2 \over {x + 1}}$$

$$f'(x) = 2{\left( {{1 \over {x + 1}}} \right)^2}$$

$$f''(x) = - 4{\left( {{1 \over {x + 1}}} \right)^3}$$

$$f'''(x) = {{12} \over {{{(x + 1)}^4}}}$$

for $$x = 1$$

$$f'''(1) = {{12} \over {{2^4}}} = {{12} \over {16}} = {3 \over 4}$$