1. ## integration 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!

2. We have $\int{\frac{dx}{e^x-1}}=\int{\frac{e^x}{(e^x-1)\cdot{e^x}}dx}$

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

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

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

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

$\int{\frac{dx}{e^x-1}}=\ln\left(1-\frac{1}{e^x}\right)+k$

3. I think you missed the sqrt

$\begin{gathered}
\int {\frac{1}
{{\sqrt {e^x - 1} }}dx} \hfill \\
u^2 = e^x - 1 \hfill \\
2udu = e^x dx \hfill \\
\end{gathered}$

$
\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}
$

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

$
\begin{gathered}
\int {\frac{{\sqrt {x^2 - 1} }}
{x}dx} \hfill \\
x^2 = u^2 + 1 \hfill \\
xdx = udu \hfill \\
\end{gathered}
$

$
\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}
$

4. Anyway if the problem would've been

Originally Posted by PaulRS
We have $\int{\frac{dx}{e^x-1}}$
then

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

Hence $\int {\frac{1}
{{e^x - 1}}\,dx} = \ln \left| {e^x - 1} \right| - x + k.$

5. Originally Posted by Krizalid
Anyway if the problem would've been

then

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

Hence $\int {\frac{1}
{{e^x - 1}}\,dx} = \ln \left| {e^x - 1} \right| - x + k.$

6. 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.

7. It's a classic problem, the answer is zero, since you're integrating an odd function on a symmetric interval.

8. Originally Posted by Krizalid
It's a classic problem, the answer is zero, since you're integrating an odd function on a symmetric interval.
Thank you for the reply but could you expand on that a bit? I'm not entirely sure what you mean by odd function.

9. Let $k=\int_{ - 1}^1 {\frac{{\sin x}}
{{1 + x^2 }}\,dx},$
substitute $u=-x,$

$k = - \int_{ - 1}^1 {\frac{{\sin u}}
{{1 + u^2 }}\,du} .$

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

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

10. Originally Posted by Krizalid
Let $k=\int_{ - 1}^1 {\frac{{\sin x}}
{{1 + x^2 }}\,dx},$
substitute $u=-x,$

$k = - \int_{ - 1}^1 {\frac{{\sin u}}
{{1 + u^2 }}\,du} .$

Hence, our original integral satisfies $k+k=0\implies k=0.$
Ok thanks a lot. That one kind of caught me off guard, as I dont remember seeing anything like that in earlier chapter assignments. I've just been doing the chapter review problems in my text.

11. Originally Posted by Krizalid
Let $k=\int_{ - 1}^1 {\frac{{\sin x}}
{{1 + x^2 }}\,dx},$
substitute $u=-x,$

$k = - \int_{ - 1}^1 {\frac{{\sin u}}
{{1 + u^2 }}\,du} .$

Hence, our original integral satisfies $k+k=0\implies k=0.$
Ok thanks a lot. That one kind of caught me off guard, as I dont remember seeing anything like that in earlier chapter assignments. It was in the chapter review problems in my text.