# Integration problem

1. Jan 23, 2009

### Melawrghk

1. The problem statement, all variables and given/known data
$$\int \frac{2e^{2t}*dt}{e^{4t}+12e^{2t}+32}$$

3. The attempt at a solution
I factored the denominator, getting (e2t+4)(e2t+8), from that I was able to express the original equation in terms of two fractions:
$$\int \frac{A}{e^{2t}+4}$$+$$\frac{B}{e^{2t}+8}$$

I then found A to be -2 and B=4.

But when I put those back into the original equations, I still have integrals that have an 'e' on the bottom and I don't know what to do next...

Any help is appreciated.

2. Jan 23, 2009

### rock.freak667

Put u=e2t, and then split into partial fractions.

3. Jan 23, 2009

### Melawrghk

More partial fractions? But I can't. u+4 doesn't factor...?

4. Jan 23, 2009

### rock.freak667

$$\int \frac{2e^{2t}}{e^{4t}+12e^{2t}+32}dt$$

u=e^2t => du/2 =e^2t dt

$$\int \frac{2e^{2t}}{e^{4t}+12e^{2t}+32}dt \equiv \frac{1}{2} \int \frac{1}{u^2+12u+32} du$$

Now do partial fractions.

5. Jan 23, 2009

### Melawrghk

I already did. It's in my procedure...

6. Jan 23, 2009

### rock.freak667

Then you'd simply have to integrate

$$\frac{1}{8} \int (\frac{1}{u+4} - \frac{1}{u+8})du$$

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