# Integration problem

## Homework Statement

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

## 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.

rock.freak667
Homework Helper
Put u=e2t, and then split into partial fractions.

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

rock.freak667
Homework Helper
More partial fractions? But I can't. u+4 doesn't factor...?

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

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.

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

rock.freak667
Homework Helper
Then you'd simply have to integrate

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