Let x 2 =t I=∫12et21+t+2t2etdtI=12∫ettet2+et2+2t2et2dtI=12∫etft+f'tdt=12ex2x2ex4+c

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$\int e^{x^{4}} x + x^{3} + 2 x^{5} e^{x^{2}} dx =$

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$\frac{1}{2} {xe}^{x^{2}} e^{x^{4}} + c$

$\frac{1}{2} x^{2} e^{x^{4}} + c$

$\frac{1}{2} e^{x^{2}} e^{x^{4}} + c$

$\frac{1}{2} x^{2} e^{x^{2}} e^{x^{4}} + c$

answer is D.

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Detailed Solution

Let $x 2 = t$
$\begin{array}{l} I = \int \frac{1}{2} e^{t^{2}} 1 + t + 2 t^{2} e^{t} dt \\ I = \frac{1}{2} \int e^{t} {te}^{t^{2}} + e^{t^{2}} + 2 t^{2} e^{t^{2}} dt \\ I = \frac{1}{2} \int e^{t} ft + f^{'} tdt = \frac{1}{2} e^{x^{2}} x^{2} e^{x^{4}} + c \end{array}$