Integral of Rational Exponential

[Jarhead1]

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

 

[skeeter]

Re: Intergral of Rational Exponential

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

$\dfrac{u'}{1+u^2}$

 

[Jarhead1]

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.

 

[MarkFL]

What skeeter is suggesting is to let:

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

And the integral becomes:

\(\displaystyle \int_1^e \frac{1}{1+u^2}\,du\)

 

[Greg]

If you haven't seen it before...

\(\displaystyle \begin{align*}y&=\arctan(x) \\
\tan(y)&=x \\
y'\sec^2(y)&=1 \\
y'&=\frac{1}{\sec^2(y)}=\frac{1}{1+\tan^2(y)}=\frac{1}{1+x^2}\end{align*}\)

 

[Jarhead1]

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!