interate[e^X^4(x+x^3+2x ^5)e^x^2]

Dear student

In question it is easier to differentiate the options and checking with qn

Differentiating option D we get

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{2}{\mathrm{x}}^2{\mathrm{e}}^{{\mathrm{x}}^2}{\mathrm{e}}^{{\mathrm{x}}^4}+\mathrm{c}\right) \\ =\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{1}{2}{\mathrm{x}}^2{\mathrm{e}}^{{\mathrm{x}}^2}{\mathrm{e}}^{{\mathrm{x}}^4}\right] \\ =\frac{1}{2}\times \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^2 . {\mathrm{e}}^{{\mathrm{x}}^2+{\mathrm{x}}^4}\right] \\ =\frac{1}{2}\times \left[{\mathrm{x}}^2 \times \left(2\mathrm{x}+4{\mathrm{x}}^3\right) .{\mathrm{e}}^{{\mathrm{x}}^2+{\mathrm{x}}^4} + {\mathrm{e}}^{{\mathrm{x}}^2+{\mathrm{x}}^4}. 2\mathrm{x} \right] \\ =\left[{\mathrm{x}}^2.{\mathrm{e}}^{{\mathrm{x}}^2+{\mathrm{x}}^4}\left(\mathrm{x}+2{\mathrm{x}}^3\right) + \mathrm{x} . {\mathrm{e}}^{{\mathrm{x}}^2+{\mathrm{x}}^4}\right] \\ ={\mathrm{x}}^3{\mathrm{e}}^{{\mathrm{x}}^2+{\mathrm{x}}^4} + 2{\mathrm{x}}^5{\mathrm{e}}^{{\mathrm{x}}^2+{\mathrm{x}}^4} + {\mathrm{xe}}^{{\mathrm{x}}^2+{\mathrm{x}}^4} \\ ={\mathrm{e}}^{{\mathrm{x}}^2+{\mathrm{x}}^4}\left[{\mathrm{x}}^3+2{\mathrm{x}}^5+\mathrm{x}\right] \\ ={\mathrm{e}}^{{\mathrm{x}}^2 }. {\mathrm{e}}^{{\mathrm{x}}^4} \left[{\mathrm{x}}^3+2{\mathrm{x}}^5+\mathrm{x}\right] \end{gathered}$$
Regards