Correct option is (2) \(\frac {3}{4}\)
\( I=\int \frac{e^x\left(x^2+1\right)}{(x+1)^2} d x=f(x) e^x+c \)
\(=\int \frac{e^x\left(x^2-1+1+1\right)}{(x+1)^2} d x \)
\(=\int e^x\left[\frac{x-1}{x+1}+\frac{2}{(x+1)^2}\right] d x\)
\(=e^x\left(\frac{x-1}{x+1}\right)+c \)
\(\therefore f(x)=\frac{x-1}{x+1} \)
\( f(x)=1-\frac{2}{x+1} \)
\( f^{\prime}(x)=2\left(\frac{1}{x+1}\right)^2 \)
\(f^{\prime \prime}(x)=-4\left(\frac{1}{x+1}\right) ^3\)
\( f^{\prime \prime \prime}(x)=\frac{12}{(x+1)^4} \)
\( \text {For } x=1, \)
\(f^{\prime \prime \prime}(1)=\frac{12}{2^4}=\frac{12}{16}=\frac{3}{4}\).
