Could use some help here:

1. The integral: 1 / sqrt( e^x - 1 ) dx

2. The integral: sqrt( x^2 - 1) / x dx

Thanks!

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- Feb 25th 2008, 12:47 PMcooliointegration problem
Could use some help here:

1. The integral: 1 / sqrt( e^x - 1 ) dx

2. The integral: sqrt( x^2 - 1) / x dx

Thanks! - Feb 25th 2008, 01:16 PMPaulRS
We have $\displaystyle \int{\frac{dx}{e^x-1}}=\int{\frac{e^x}{(e^x-1)\cdot{e^x}}dx}$

Let $\displaystyle u=e^x$ then $\displaystyle \frac{du}{dx}=e^x$

We get: $\displaystyle \int{\frac{dx}{e^x-1}}=\int{\frac{du}{(u-1)\cdot{u}}}$

See that $\displaystyle \frac{1}{(u-1)\cdot{u}}=\frac{1}{u-1}-\frac{1}{u}$

So: $\displaystyle \int{\frac{du}{(u-1)\cdot{u}}}=\int{\frac{du}{u-1}}-\int{\frac{du}{u}}=\ln(1-1/u)+k$

$\displaystyle \int{\frac{dx}{e^x-1}}=\ln\left(1-\frac{1}{e^x}\right)+k$ - Feb 25th 2008, 01:28 PMPeritus
I think you missed the sqrt

$\displaystyle \begin{gathered}

\int {\frac{1}

{{\sqrt {e^x - 1} }}dx} \hfill \\

u^2 = e^x - 1 \hfill \\

2udu = e^x dx \hfill \\

\end{gathered}$

$\displaystyle

\int {\frac{1}

{u}\frac{{2u}}

{{u^2 + 1}}du} = 2\int {\frac{1}

{{1 + u^2 }}du = 2\arctan u + C = 2\arctan \sqrt {e^x - 1} + C}

$

-------------------------------------------------------------------------

$\displaystyle

\begin{gathered}

\int {\frac{{\sqrt {x^2 - 1} }}

{x}dx} \hfill \\

x^2 = u^2 + 1 \hfill \\

xdx = udu \hfill \\

\end{gathered}

$

$\displaystyle

\int {\frac{{u^2 }}

{{u^2 + 1}}du = } \int {1 - \frac{1}

{{u^2 + 1}}du = u - \arctan u + C = \sqrt {x^2 - 1} - \arctan \sqrt {x^2 - 1} + C}

$ - Feb 25th 2008, 01:51 PMKrizalid
Anyway if the problem would've been

then

$\displaystyle \int {\frac{1}

{{e^x - 1}}\,dx} = \int {\frac{{e^x - \left( {e^x - 1} \right)}}

{{e^x - 1}}\,dx} = \int {\frac{{\left( {e^x - 1} \right)'}}

{{e^x - 1}}\,dx} - \int {dx}.$

Hence $\displaystyle \int {\frac{1}

{{e^x - 1}}\,dx} = \ln \left| {e^x - 1} \right| - x + k.$ - Feb 25th 2008, 02:41 PMPeritus
- Feb 25th 2008, 02:51 PMcoolio
Thanks a for the help everyone!

I came across another one that has stumped me:

The integral from -1 to 1: sinx / 1 + x^2 dx

I see 1 / 1+x^2 as the derivative of arctan, but I'm not sure if that is relevant. The only this I could think of was to integrate by parts, but after attempting that, I seem to just be going in circles. - Feb 25th 2008, 02:52 PMKrizalid
It's a classic problem, the answer is zero, since you're integrating an odd function on a symmetric interval.

- Feb 25th 2008, 02:57 PMcoolio
- Feb 25th 2008, 03:01 PMKrizalid
Let $\displaystyle k=\int_{ - 1}^1 {\frac{{\sin x}}

{{1 + x^2 }}\,dx},$ substitute $\displaystyle u=-x,$

$\displaystyle k = - \int_{ - 1}^1 {\frac{{\sin u}}

{{1 + u^2 }}\,du} .$

Hence, our original integral satisfies $\displaystyle k+k=0\implies k=0.$

(Ahh, I just saw you edited your original question: an odd function $\displaystyle f(x)$ satisfies $\displaystyle f(-x)=-f(x).$ When having integrals whose function is odd besides if they're taken on a symmetric interval $\displaystyle [-a,a]$ the answer is zero.) - Feb 25th 2008, 03:05 PMcoolio
- Feb 25th 2008, 03:06 PMcoolio