rational substiution integral

**akhayoon** · January 24th 2008, 02:40 PM

$\int \frac{dx}{\sqrt{1+e^{x}}}$

$u^{2}=1+e^{x}$

[tex] 2u du=e^{x} dx [tex]

I finally get to

$\frac{2}{3} (\int \frac{1}{u-1}du+ \int \frac{u+1}{u^{2}+u+1}du)$

so if this is right

can somebody help me with hte second part of the last integral

and if it's wrong somekind of solution

**Jhevon** · January 24th 2008, 02:53 PM

Originally Posted by akhayoon

$\int \frac{dx}{\sqrt{1+e^{x}}}$

$u^{2}=1+e^{x}$

[tex] 2u du=e^{x} dx [tex]

I finally get to

$\frac{2}{3} (\int \frac{1}{u-1}du+ \int \frac{u+1}{u^{2}+u+1}du)$

so if this is right

can somebody help me with hte second part of the last integral

and if it's wrong somekind of solution

i get something different for the integral. by the way, i don't like the substitution you made because $\sqrt{u^2} = |u| \ne u$ , but anyway, you'd get something like $2 \int \frac {du}{u^2 - 1}$ if we say $u > 0$

an alternative is to use two substitutions. first $u = 1 + e^x$ and then $t = \sqrt{u}$ and you'll get a similar integral to the one above

if you're still interested, completing the square coupled with substitution can take care of your original integral

**Soroban** · January 24th 2008, 08:02 PM

Hello, akhayoon!

Your substitution should have come out like Jhevon's . . .

$\int \frac{dx}{\sqrt{1+e^{x}}}$

Let $u = \sqrt{1+e^x}\quad\Rightarrow\quad e^x \:=\:u^2-1\quad\Rightarrow\quad x \:=\:\ln(u^2-1) \quad\Rightarrow\quad dx \:=\:\frac{2u\,du}{u^2-1}$

Substitute: . $\int\frac{1}{u}\cdot\frac{2u\,du}{u^2-1} \;=\;2\int\frac{du}{u^2-1} \;=\;\ln\left|\frac{u-1}{u+1}\right| + C$

Back-substitute: . $\boxed{\ln\left|\frac{\sqrt{1+e^x} - 1}{\sqrt{1+e^x} + 1}\right| + C}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here's another method . . . it's either brilliant or incredibly stupid.

Let $e^x \:=\:\tan^2\theta\quad\Rightarrow\quad x \:=\:2\ln|\tan\theta|\quad\Rightarrow\quad dx \:=\:\frac{2\sec^2\!\theta}{\tan\theta}\,d\theta$

Substitute: . $\int\frac{1}{\sec\theta}\cdot\frac{2\sec^2\!\theta }{\tan\theta}\,d\theta \;=\;2\int\csc\theta\,d\theta \;=\;2\!\cdot\!\ln|\csc\theta - \cot\theta| + C$

Back-substitute

We have: . $\tan\theta \:=\:e^{\frac{x}{2}}$

This is a right triangle with: . $opp = e^{\frac{x}{2}},\;adj = 1$
Hence: . $hyp = \sqrt{1+e^x} \quad\Rightarrow\quad\csc\theta = \frac{\sqrt{1+e^x}}{e^{\frac{x}{2}}},\;\cot\theta = \frac{1}{e^{\frac{x}{2}}}$

So we have: . $2\!\cdot\!\ln\left|\frac{\sqrt{1+e^x}}{e^{\frac{x} {2}}} - \frac{1}{e^{\frac{x}{2}}}\right| + C \;=\;2\!\cdot\!\ln\left|\frac{\sqrt{1+e^x} - 1}{e^{\frac{x}{2}}}\right| + C$

. . $= \;2\ln\left|\sqrt{1+e^x} - 1\right| - 2\ln\left|e^{\frac{x}{2}}\right| + C \;=\;2\!\cdot\!\ln\left|\sqrt{1+e^x} - 1\right| - \ln\left(e^{\frac{x}{2}}\right)^2 + C$

. . $= \;2\!\cdot\!\ln\left|\sqrt{1+e^x} - 1\right| - \ln(e^x) + C \;=\;\boxed{2\!\cdot\!\ln\left|\sqrt{1+e^x} - 1\right| - x + C}<br />$

You mission (should you decide to accept it)
. . is to prove that my two answers are equivalent.
.

**Jhevon** · January 25th 2008, 11:50 AM

Originally Posted by Soroban

Here's another method . . . it's either brilliant or incredibly stupid.

i say brilliant! so crazy, it's brilliant! ...though not for the faint of heart, or those new to the world of integrals. how did you discover such a substitution? was the integral given in a "special form"?

You mission (should you decide to accept it)
. . is to prove that my two answers are equivalent.
.

later

**Krizalid** · January 25th 2008, 12:01 PM

One can play with trig. sub. as many times as necessary.

700% personal opinion, I don't like trig. substitutions, some times it turns the integrand into a mess, but it depends of the problem.

I prefer workin' with algebra, you can get faster solutions, especially the reciprocal substitution, which can turn a nasty integral, into a cute one. (Not always, of course.)

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