$\int \frac{\left(x^2+1\right)e^x}{(x+1{)}^2}dx=f\left(x\right)e^x+C$, Where $\mathrm{C}$ is a constant, then $\frac{d^{3}f}{d{x}^3}$ at $\mathrm{x}=1$ is equal to :

$\begin{array}{l}\int \left(\frac{x^2+1}{{(x+1)}^2}\right)e^x·dx\\ \\ =\int \left(\frac{x^2-1+2}{{(x+1)}^2}\right)e^{x}dx\\ \\ =\int \left(\frac{x-1}{x+1}+\frac{2}{{(x+1)}^2}\right)e^{x}dx\\ \\ =\int \left(f\left(x\right)+f^{\text{'}}\left(x\right)\right)e^{x}dx\\ \\ =f\left(x\right)e^x+c\\ \\ Where f\left(x\right)=\frac{x-1}{x+1}\\ \\ f^{\text{'}}\left(x\right)=\frac{2}{{(x+1)}^2}\\ \\ \begin{array}{l}{f}^{\text{'}\text{'}}\left(x\right)=\frac{-4}{{(x+1)}^3}\\ \\ f\text{'}\text{'}\text{'}\left(x\right)=\frac{12}{{(x+1)}^4}\\ \\ f^{\text{'}\text{'}\text{'}}\left(1\right)=\frac{12}{16}\\ \\ =\frac{3}{4}\end{array}\end{array}$