MHB 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.$$\int_{0}^{1} e^{x}/(1 + e^{2x}) \,dx$$

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. $$\int e^{x}dx$$ is $$e^{x}+C$$ but that doesn't help with $$1/ (1 + e^{2x})$$. I tried u sub of $$u = 1 + e^{2x}$$ but $$du/2e^{2x} = dx $$ doesn't help. I end up with this integral with a u sub and $$e^{-x}$$ .

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

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:

$$\int \frac{1}{1 + {x}^{2}} dx = arctan$$

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

In du notation: $$\int \frac{1}{1 + {u}^{2}} du$$

The problem is getting the $$e^{2x} $$ to replace the $$e^{x}$$ in this case... ? Derivative of $$e^{2x}$$ is $$2e^{2x}$$ which doesn't replace the $$e^{x}$$ unless I am doing it wrong.

 

[MarkFL]

What skeeter is suggesting is to let:

$$u=e^x\implies du=e^x\,dx$$

And the integral becomes:

$$\int_1^e \frac{1}{1+u^2}\,du$$

 

[Greg]

If you haven't seen it before...

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

 

[Jarhead1]

Ah ok. Was having trouble visualizing the exponents.

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

I was stuck on having to replace the $$e^{x}$$ with a $$e^{2x}$$ whole.

Thanks!