# Integration Problem: Solving an Equation with e

• Melawrghk
In summary, the original integral can be solved by first factoring the denominator and then expressing it as a sum of two fractions. After finding the values for A and B, the integral can be rewritten in terms of u=e^2t and solved through partial fractions. This leads to a final integral that can be easily integrated.
Melawrghk

## 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.

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

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

Melawrghk said:
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...

Then you'd simply have to integrate

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

## 1. What is the general process for solving an equation with e?

The general process for solving an equation with e is to first isolate the term containing e on one side of the equation. Then, take the natural logarithm of both sides to eliminate the e. Finally, solve for the remaining variable.

## 2. How do I know when to use the natural logarithm when solving an equation with e?

You should use the natural logarithm when the equation contains e as an exponent or when the equation can be rewritten to have e as an exponent. This is because taking the natural logarithm will eliminate the e, making it easier to solve for the variable.

## 3. Can an equation with e have multiple solutions?

Yes, an equation with e can have multiple solutions. This is because the natural logarithm is a multivalued function, meaning it can have multiple solutions for a given input. Therefore, when solving an equation with e, you should always check for extraneous solutions.

## 4. Can e be solved for in an equation?

No, e cannot be solved for in an equation. This is because e is a constant, representing a specific numerical value (approximately 2.71828). It is not a variable that can be solved for.

## 5. Are there any specific rules for solving an equation with e?

Yes, there are a few rules to keep in mind when solving an equation with e. First, the natural logarithm of a negative number is undefined, so any solutions that result in a negative number under the natural logarithm should be discarded. Additionally, when taking the natural logarithm of both sides, the result should be simplified to the form "x = ...".

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