# Integral + brainfart

1. Aug 4, 2009

### James889

Hai,

I have the following Integral i can't get right:
$$\int_{-8}^8\frac{(6e^{4x}-2)^2}{e^{4x}}$$

After i squared the bracket i end up with$$~~\frac{36e^{8x}-24e^{8x}+4}{e^{4x}}$$

So, after dividing thru by $$e^{4x}$$ i have:

$$36e^{4x}-24e^{4x}+\frac{4}{e^{4x}}$$

Integrating gives:
$$9e^{4x} - 6e^{4x} +4~ln(e^{4x})$$

I must be doing something wrong because i end up with the wrong answer =/
But what?

2. Aug 4, 2009

### zcd

$$\int 36e^{4x}-24e^{4x}+\frac{4}{e^{4x}}\,dx=\int 36e^{4x}-24e^{4x}+4e^{-4x}}\,dx$$
You skipped the u substitution for the last term.

3. Aug 4, 2009

### rl.bhat

The third integration becomes
Int(4*e^-4x) = - e^-4x.

4. Aug 4, 2009

### Pengwuino

The second term in line 2 should be $$- 24e^{4x}$$. Also, the integral of $$4e^{ - 4x}$$ is not $$4\ln (e^{4x} )$$