# 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?

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What is the $f$, and where did it go? It seems to have magically disappeared in the middle of your working. –  user22805 Jul 21 '12 at 11:06
It will be highly unlikely anyone will understand your question, let alone solve it, without you telling us what is that "f" appearing there... –  DonAntonio Jul 21 '12 at 11:56
In the title, there's a 2 in the denominator; in the first line of the body, it's $e^{2x}$; in the second line, it's $e^{2x}$ on the left, and 2 on the right. I suggest you heed the comments, and figure out exactly what it is that you want to ask, and then edit your question accordingly. –  Gerry Myerson Jul 21 '12 at 12:29
And if it's $e^{2x}$ in the denominator, then you have made the mistake of doing $\int(g/h)=(\int g)/(\int h)$. –  Gerry Myerson Jul 21 '12 at 12:32
Both the "f" and the "2" (in the title) were typos, thanks for letting me know. –  Quispiam Jul 21 '12 at 13:06
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$$\int_{-6}^6 \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x= \int_{-6}^6 \frac{16e^{4x} + 16e^{2x}+ 4}{e^{2x}} \, \mathrm{d} x= \int_{-6}^6 16e^{2x} + 16+ 4e^{-2x} \, \mathrm{d} x= \left[ 8e^{2x} + 16x-2e^{-2x} \right]_{-6}^6= 8(e^{12}-e^{-12}) + 16\cdot 12 -2(e^{-12}-e^{12})= 192+ 10 e^{12}-10 e^{-12}$$

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.

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+1 Certainly the most direct approach! –  alex.jordan Jul 21 '12 at 18:20

$$I:=\int_{-6}^6 \frac{(4e^{2x} + 2)^2}{e^{2x}}\ dx$$

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

$$I \approx 1.628\times10^6$$

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You forgot to change the limits of the integral to $e^{\pm 6}$ –  Belgi Jul 21 '12 at 13:48
After doing the change of variables, shouldn't it be $u^2$ instead of $e^{2u}$ in the numerator? –  Javier Badia Jul 21 '12 at 15:46
Yes, you are right. I made quite a few mistakes when I wrote this up right after I got out of bed, sorry. –  Joe Jul 21 '12 at 16:08
+1. An observation: subbing $u=e^{2x}$ leaves a "tidier" rational function that is merely quadratic over quadratic. –  alex.jordan Jul 21 '12 at 18:17
Yup. Turns out there are many ways to attack this integral, which is pretty common for textbook exercises (which I imagine is where this came from). –  Joe Jul 21 '12 at 19:05

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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Note that this answer is also equal to $4 (48 + 5 \sinh(12))$ –  Matt Groff Jul 21 '12 at 17:30
$$\int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x = \left[ \frac{(4e^{4x} + 8e^{2x} + 4x)2}{e^{2x}} \right]_{-6}^6$$
The 2 outside the parenthesis came from integrating the denominator. As I wrote in a comment, the "rule" $\int(f/g)=(\int f)/(\int g)$ is being used. –  Gerry Myerson Jul 22 '12 at 7:07