Explanation

(d)
Let us assume $f(x) = \frac{e^{x^{2}}}{2}$
$\therefore\ \int_{}^{}{x.\frac{e^{x^{2}}}{2}dx = \int_{}^{}{2x.\frac{e^{x^{2}}}{2}dx}}$
Put $x^{2} = t\ \ \Rightarrow \ 2x\ dx = dt\$
$\therefore\ \int_{}^{}\frac{e^{t}}{4}dt = \frac{e^{t}}{4} + c = \frac{1}{2}\left( \frac{e^{x^{2}}}{2} \right) + c$
Hence, $\int_{}^{}xf(x)dx = \frac{1}{2}f(x) + c$, when $f(x) = \frac{e^{x^{2}}}{2}$