# Numerical integration

#### Albert

##### Well-known member
$2.37=\frac{1}{\sqrt{6}} \int_{0}^{x} \sqrt{\frac{e^x}{e^x-1}}dx$

please find x to three decimal point

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#### MarkFL

Staff member
Using technology and some guesswork, I find:

$\displaystyle \frac{1}{\sqrt{6}}\int_0^{4.425}\sqrt{\frac{e^t}{e^t-1}}\,dt=2.37$

$x=4.425$

#### Albert

##### Well-known member
I get x $\approx$ 4.428

#### MarkFL

Staff member
wolframalpha.com gave the result I cited as being exact, but my TI-89 gives:

$x\approx4.42501043622$

#### Bacterius

##### Well-known member
MHB Math Helper
wolframalpha.com gave the result I cited as being exact, but my TI-89 gives:

$x\approx4.42501043622$
The TI is probably having floating-point accuracy problems - it's not exactly a trivial approximation - I doubt W|A is wrong. I would use Mathematica to confirm but I don't have it installed right now

#### MarkFL

Staff member
W|A also says the value my TI-89 gave results in the same exact value as well.

#### Jester

##### Well-known member
MHB Math Helper
Do you really want the upper limit and dummy variable (the integration variable) to both be $x$?

#### Jester

##### Well-known member
MHB Math Helper
If the integration variable is in fact $t$ (like MarkFL notes), then we can solve exactly for $x$ giving

$x = 2 \ln \dfrac{k^2+1}{2k}$ where $k = e^{2.37\sqrt{6}/2}$.

#### MarkFL

Staff member
If the integration variable is in fact $t$ (like MarkFL notes), then we can solve exactly for $x$ giving

$x = 2 \ln \dfrac{k^2+1}{2k}$ where $k = e^{2.37\sqrt{6}/2}$.
Nice!

Did you find this by directly computing the improper integral, or did you exploit the FTOC in some other way?

#### Jester

##### Well-known member
MHB Math Helper
Nice!

Did you find this by directly computing the improper integral, or did you exploit the FTOC in some other way?
I integrated directly. Once you let $u = e^{t/2}$, you get an integral very manageable.