∫ x 2 + 1 e x x + 1 2 d x = f x e x + C , where C is a constant, then d 3 f d x 3 at x = 1 is equal to

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

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

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

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

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

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

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

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

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

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

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