integral of (2e^x - 2e^-x)/(e^x + e^-x) dx

[Guest]

I solved then put into my TI89 and got a very different answer. Can someone help?

integral of (2e^x - 2e^-x)/(e^x + e^-x) dx

My work:

u = e^x + e^-x, du = e^x + -1e^-x dx

2 integral 1/u^2 du

2* -1/u

-2/(e^x+e^-x)

Is this correct? And, if not, where did I go wrong?

Thanks

 

[pka]

\(\displaystyle \L
\begin{array}{rc}
\int {\frac{{2e^x - 2e^{ - x} }}{{e^x + e^{ - e} }}} & = & \int {\frac{{2e^{2x} - 2}}{{e^{2x} + 1}}} \\
& = & \int {\frac{{2e^{2x} }}{{e^{2x} + 1}} - \int {\frac{2}{{e^{2x} + 1}}} } \\
\end{array}\)

 

[galactus]

What you have is the equivalent of \(\displaystyle 2tanh(x)\)

\(\displaystyle \L\\2\int{tanh(x)}dx\)

=\(\displaystyle \L\\2\int\frac{sinh(x)}{cosh(x)}dx\)

Let \(\displaystyle u=cosh(x)\;\ and du=sinh(x)dx\)

You get:

\(\displaystyle \L\\ln(cosh(x))+C\)

 

[Guest]

Am I wrong to say you can not have two answers for the same integral?

 

[pka]

The latter answer is a better way to get to a more compact answer.
The two are actually equivlant.

 

[soroban]

Re: integral

Hello, ezrajoelmicah!

I solved then put into my TI89 and got a very different answer.

Of course you did . . . Your answer is wrong.

\(\displaystyle \L\int \frac{2e^x\,-\,2e^{-x}}{e^x\,+\,e^{-x}}\,dx\)

My work:

\(\displaystyle u\:=\:e^x\,+\,e^{-x}\;\;\Rightarrow\;\; du\:=\:\left(e^x\,-\,e^{-x}\right)\,dx\)

Substitute: \(\displaystyle \L\:2\int\frac{e^x\,-\,e^{-x}}{e^x\,+\,e^{-x}}\,dx\;=\;2\int\frac{du}{u}\)

. . . . . . .\(\displaystyle \L= \;2\cdot\ln(e^x\,+\,e^{-x})\,+\,C\)

 

[pka]

Well MathCad give a different and equivalent answer:
\(\displaystyle \L
2\ln \left( {e^x } \right) - \ln \left( {1 + e^{2x} } \right)\)