<s>[INST] 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?
 [/INST] You had these steps ok:
$$
\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x = 
\int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x
$$
After that, there are a number of choices.  It looks like you forgot to integrate the solution.
You could do this:
$$\int_{-6}^6 {\frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} dx}$$
$$= \int_{-6}^6 { \left( 16e^{2x} + 16 + 4e^{-2x} \right) dx}$$
$$= \left[ { 8e^{2x} + 16x - 2e^{-2x} } \right]_{-6}^6$$
The integration is directly above.  Plugging in the values then gives:
$$ \left( 8e^{2(6)} + 16(6) - 2e^{-2(6)} \right) -  \left( 8e^{2(-6)} + 16(-6) - 2e^{-2(-6)} \right) $$
$$= \left( 8e^{12} + 96 - 2e^{-12} \right) -  \left( 8e^{-12} -96 - 2e^{12} \right) $$
$$= 10e^{12} + 192 - 10e^{-12} $$
$$\approx 1.62774*10^6$$
To get the hyperbolic sine ($\sinh$), note that 
$$  \sinh(x)  = \frac{ e^{x}  - e^{-2x} } {2}$$
$$  \sinh(12) = \frac{ e^{12} - e^{-12} } {2}$$
$$20\sinh(12) = 10 \left( e^{12} - e^{-12} \right)$$
So we have
$$  10e^{12} - 10e^{-12} + 192 $$
$$= 20\sinh(12) + 192 $$
$$= 4 \left( 5 \sinh(12) + 48 \right)$$
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