Compute integral $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ I want to solve $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ but I get the wrong results:
$$
\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x = 
\int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x
$$
$$
= \left[ \frac{(4e^{4x} + 8e^{2x} + 4x)2}{e^{2x}} \right]_{-6}^6 =
\left[ \frac{8e^{4x} + 16e^{2x} + 8x}{e^{2x}} \right]_{-6}^6
$$
$$
= (\frac{8e^{24} + 16e^{12} + 48}{e^{12}}) - (\frac{8e^{-24} + 16e^{-12} - 48}{e^{-12}})
$$
$$
= e^{-12}(8e^{24} + 16e^{12} + 48) - e^{12}(8e^{-24} + 16e^{-12} - 48)
$$
$$
= 8e^{12} + 16 + 48e^{-12} - (8e^{-12} + 16 - 48e^{12})
$$
$$
= 8e^{12} + 16 + 48e^{-12} - 8e^{-12} - 16 + 48e^{12})
$$
$$
= 56e^{12} + 56e^{-12}
$$
Where am I going wrong?

$$
\int_{-6}^6 \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x=
\int_{-6}^6 \frac{16e^{4x} + 16e^{2x}+ 4}{e^{2x}} \, \mathrm{d} x=
\int_{-6}^6 16e^{2x} + 16+ 4e^{-2x} \, \mathrm{d} x=
\left[ 8e^{2x} + 16x-2e^{-2x} \right]_{-6}^6=
8(e^{12}-e^{-12}) + 16\cdot 12 -2(e^{-12}-e^{12})=
192+ 10 e^{12}-10 e^{-12}
$$
You can check both indefinite
and definite
integral at WolframAlpha.

I am not sure where is mistake in your solution (since I do not understand what exactly you have done), but most probably you have used $\int \frac{f(x)}{g(x)} \, \mathrm{d} x = \frac{\int f(x) \, \mathrm{d} x}{\int g(x)\, \mathrm{d} x}$, as suggested by Gerry's comment. This formula is incorrect.
