QJEEM22Mafbeb: JEE Main 2022 Mathematics - Class 12 JEEM22Mafbeb Solution: If , where C is a constant, then at x = 1 is equal to :

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

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

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

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

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

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

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

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

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

for $x = 1$

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