If [ (x^2 + 1)e^x (x + 1)^2dx = f(x)e^x + C ], where C is a constant, then [d^3f dx^3] at …

Question

If $$\int {{{({x^2} + 1){e^x}} \over {{{(x + 1)}^2}}}dx = f(x){e^x} + C} $$, where C is a constant, then $${{{d^3}f} \over {d{x^3}}}$$ at x = 1 is equal to :

Accepted Answer

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

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

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

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

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

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

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

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

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

for $x = 1$

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

Solutions · 1