Explanation:
(D) : Given, \(\quad I=\int_{0}^{2} \frac{x^{2}}{\left(x^{3}+1\right)^{2}} d x\)
\(\begin{array}{ll}\text { Let, } x^{3}+1=t \\ 3 x^{2} d x=d t\end{array}\)
\(\mathrm{x}^{2} \mathrm{dx}=\frac{1}{3} \mathrm{dt}\)
When, \(x=0, t=1\)
\(\mathrm{x}=2, \mathrm{t}=9\)
Now, \(\quad \mathrm{I}=\frac{1}{3} \int_{1}^{9} \frac{1}{\mathrm{t}^{2}} \mathrm{dt}\)
\(=\frac{-1}{3}\left[\frac{1}{t}\right]_{1}^{9}=\frac{-1}{3}\left[\frac{1}{9}-1\right]\)
\(=\frac{-1}{3} \times \frac{-8}{9}=\frac{8}{27}\)