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?
 A: $$
\int_{-6}^6 \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x=
\int_{-6}^6 \frac{16e^{4x} + 16e^{2x}+ 4}{e^{2x}} \, \mathrm{d} x=
\int_{-6}^6 16e^{2x} + 16+ 4e^{-2x} \, \mathrm{d} x=
\left[ 8e^{2x} + 16x-2e^{-2x} \right]_{-6}^6=
8(e^{12}-e^{-12}) + 16\cdot 12 -2(e^{-12}-e^{12})=
192+ 10 e^{12}-10 e^{-12}
$$
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.
A: $$ I:=\int_{-6}^6  \frac{(4e^{2x} + 2)^2}{e^{2x}}\ dx$$
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
$$I \approx 1.628\times10^6$$
A: 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
$$  10e^{12} - 10e^{-12} + 192 $$
$$= 20\sinh(12) + 192 $$
$$= 4 \left( 5 \sinh(12) + 48 \right)$$
A: As for the error in your work, I see a problem in the following step:
$$
\int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x
= \left[ \frac{(4e^{4x} + 8e^{2x} + 4x)2}{e^{2x}} \right]_{-6}^6$$
The denominator is not a constant, so you cannot do the integration like this. I would suggest dividing the numerator by the denominator. This amounts to the substitution which Joe suggests, but seems less complicated in my opinion.
Also, the 2 outside the parenteses in the numerator is incorrect.
