What is the surface area of the solid created by revolving f(x) =e^(4x)-x^2e^(2x) , x in [1,2] around the x axis?

Given that

$f \left(x\right) = y = {e}^{4 x} - {x}^{2} {e}^{2 x}$

Volume of solid obtained by revolving the above curve about the x-axis between $x = 1$ & $x = 2$

$V = \setminus {\int}_{1}^{2} {y}^{2} \setminus \mathrm{dx}$

$= \setminus {\int}_{1}^{2} {\left({e}^{4 x} - {x}^{2} {e}^{2 x}\right)}^{2} \setminus \mathrm{dx}$

$= \setminus {\int}_{1}^{2} \left({e}^{8 x} + {x}^{4} {e}^{4 x} - 2 {x}^{2} {e}^{6 x}\right) \setminus \mathrm{dx}$

$= \setminus {\int}_{1}^{2} \left({e}^{8 x} + {x}^{4} {e}^{4 x} - 2 {x}^{2} {e}^{6 x}\right) \setminus \mathrm{dx}$
$= {\left[{e}^{8 x} / 8 + {e}^{4 x} \left({x}^{4} / 4 - {x}^{3} / 4 + \frac{3 {x}^{2}}{16} - \frac{3 x}{32} + \frac{3}{128}\right) - {e}^{6 x} \left({x}^{2} / 3 - \frac{x}{9} + \frac{1}{54}\right)\right]}_{1}^{2}$

$= 934337.8548$

The surface area is given as

$\setminus {\int}_{1}^{2} 2 \setminus \pi y \setminus \sqrt{1 + {\left(y '\right)}^{2}} \setminus \mathrm{dx}$

$= \setminus {\int}_{1}^{2} 2 \setminus \pi \left({e}^{4 x} - {x}^{2} {e}^{2 x}\right) \setminus \sqrt{1 + {\left({e}^{4 x} - 2 x {e}^{2 x} \left(x + 1\right)\right)}^{2}} \setminus \mathrm{dx}$

Solving above integral is much more complicated which can't be solved by elementary method