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?

$$ I:=\int_{-6}^6  \frac{(4e^{2x} + 2)^2}{e^{2x}}\ dx$$
Let $u=e^x, du = e^x \ dx$, leaving us with:
$$\int_{e^{-6}}^{e^{6}} \frac{\left( 4u^2 + 2 \right)^2}{u^3} \ du$$
Expand the numerator to get
$$\int_{e^{-6}}^{e^{6}} \frac{16u^4 + 16u^2 + 4}{u^3} \ du$$
Since the highest power in the numerator is greater than the highest power in the denominator, we have to do some long division. Upon dividing, you get:
$$\int_{e^{-6}}^{e^{6}} \frac{4}{u^3} + 16u + \frac{16}{u} \ du$$
Integrate to get:
$$8u^2-\frac{2}{u^2} + 16 \ln |u|$$
Back-substitute $u=e^x$ to get
$$8e^{2x} - 2e^{-2x} + 16 \ln|e^{x}|$$
Since $e^x$ is strictly increasing, we can drop the absolute value. Also, recall that $\ln{e^x} = x$, so you can simplify a bit.
$$8e^{2x} + 16x - 2e^{-2x}$$
Now, simply evaluate at your endpoints to find that
$$I \approx 1.628\times10^6$$
