Integral of Rational Exponential

In summary, the problem is that the e^{2x} is not being replaced by the e^{x} in the equation, so the integral cannot be solved.
  • #1
Jarhead1
3
0
Hi,

I'm new to this forum. This semester I took Calculus I and just took the final yesterday. There were a few questions that were unexpected that I didn't know how to handle. This integral has got me stumped.\(\displaystyle \int_{0}^{1} e^{x}/(1 + e^{2x}) \,dx\)

The techniques I know at this point include u substitution and the table of integral rules which I'm sure is limited at this point. \(\displaystyle \int e^{x}dx\) is \(\displaystyle e^{x}+C\) but that doesn't help with \(\displaystyle 1/ (1 + e^{2x})\). I tried u sub of \(\displaystyle u = 1 + e^{2x}\) but \(\displaystyle du/2e^{2x} = dx \) doesn't help. I end up with this integral with a u sub and \(\displaystyle e^{-x}\) .

\(\displaystyle 1/2 \int_{0}^{1} 1/u \cdot e^{-x} \,du\) Note: \(\displaystyle e^{x}/e^{2x} = e^{-x}\)

Maybe there is a technique we haven't learned yet or I missed something.

Thanks in advance
 
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  • #2
Re: Intergral of Rational Exponential

... does this form of a function's derivative look familiar?

$\dfrac{u'}{1+u^2}$
 
  • #3
Re: Intergral of Rational Exponential

No I have not seen that exact form. It looks similar to the anti-derivative of arctan:

\(\displaystyle \int \frac{1}{1 + {x}^{2}} dx = arctan\)

Not sure where the \(\displaystyle {u}^{'} \) comes from unless you are referring to the du from the u substitution in prime notation.

In du notation: \(\displaystyle \int \frac{1}{1 + {u}^{2}} du\)

The problem is getting the \(\displaystyle e^{2x} \) to replace the \(\displaystyle e^{x}\) in this case... ? Derivative of \(\displaystyle e^{2x}\) is \(\displaystyle 2e^{2x}\) which doesn't replace the \(\displaystyle e^{x}\) unless I am doing it wrong.
 
  • #4
What skeeter is suggesting is to let:

\(\displaystyle u=e^x\implies du=e^x\,dx\)

And the integral becomes:

\(\displaystyle \int_1^e \frac{1}{1+u^2}\,du\)
 
  • #5
If you haven't seen it before...

\(\displaystyle \begin{align*}y&=\arctan(x) \\
\tan(y)&=x \\
y'\sec^2(y)&=1 \\
y'&=\frac{1}{\sec^2(y)}=\frac{1}{1+\tan^2(y)}=\frac{1}{1+x^2}\end{align*}\)
 
  • #6
Ah ok. Was having trouble visualizing the exponents.

So rewrite \(\displaystyle e^{2x}\) as \(\displaystyle (e^{x})^{2}\) then replace the \(\displaystyle e^{x}\) with u.

I was stuck on having to replace the \(\displaystyle e^{x}\) with a \(\displaystyle e^{2x}\) whole.

Thanks!
 

FAQ: Integral of Rational Exponential

1. What is the definition of the integral of a rational exponential function?

The integral of a rational exponential function is a mathematical concept that represents the area under the curve of a rational exponential function. It is denoted by the symbol ∫ and is used to find the total value of a function within a given interval.

2. How is the integral of a rational exponential function calculated?

The integral of a rational exponential function can be calculated using various methods, such as substitution, integration by parts, or partial fraction decomposition. The specific method used depends on the complexity of the function.

3. What is the significance of the integral of a rational exponential function?

The integral of a rational exponential function has many applications in mathematics, physics, and engineering. It is used to calculate areas, volumes, and other quantities that are represented by a rational exponential function.

4. Can the integral of a rational exponential function be negative?

Yes, the integral of a rational exponential function can be negative. This can happen when the function has negative values within the given interval, resulting in a negative area under the curve.

5. Is there a connection between the integral of a rational exponential function and the derivative of the function?

Yes, there is a fundamental connection between the integral and derivative of a rational exponential function. The fundamental theorem of calculus states that the derivative of the integral of a function is equal to the original function. Therefore, the derivative of the integral of a rational exponential function will be the original function itself.

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