∫ ( x ^ 2 + 1) e ^ x (x + 1) ^ 2 dx = f(x) e ^ x + C , where C is a...

$∫ \dfrac{\left({x}^{2} + 1\right){e}^{x}dx}{{\left(x + 1\right)}^{2}} = f\left(x\right){e}^{x} + C$

$∫ {e}^{x}\left(\dfrac{{x}^{2} - 1}{{\left(x + 1\right)}^{2}} + \dfrac{2}{{\left(x + 1\right)}^{2}}\right)dx = f\left(x\right){e}^{x} + C$

$∫ {e}^{x}\left(\dfrac{x - 1}{x + 1} + \dfrac{2}{{\left(x + 1\right)}^{2}}\right)dx = f\left(x\right){e}^{x} + C$

We know that $∫ {e}^{x}\left(f\left(x\right) + {f}^{'}\left(x\right)\right) = {e}^{x}f\left(x\right) + c$

Here $f\left(x\right) = \dfrac{x - 1}{x + 1} \text{ and } {f}^{'}\left(x\right) = \dfrac{2}{{\left(x + 1\right)}^{2}}$

So $∫ {e}^{x}\left(\left(\dfrac{x - 1}{x + 1}\right) + \dfrac{2}{( x + 1{)}^{2}}\right)dx = f\left(x\right){e}^{x} + C$

$⇒ {e}^{x}\left(\dfrac{x - 1}{x + 1}\right) + C = {e}^{x}f\left(x\right) + C$

On comparing both sides we get $f\left(x\right) = \dfrac{x - 1}{x + 1}$

So ${f}^{'}\left(x\right) = \dfrac{2}{{\left(x + 1\right)}^{2}} \text{ and } {f}^{"}\left(x\right) = \dfrac{- 4}{{\left(x + 1\right)}^{3}}$

${f}^{'''}\left(x\right) = \dfrac{12}{{\left(x + 1\right)}^{4}}$ ,

Now ${f}^{'''}\left(1\right) = \dfrac{12}{{\left(1 + 1\right)}^{4}} = \dfrac{12}{16} = \dfrac{3}{4}$