Hi
Evaluate:
$\displaystyle \int \frac{1}{\sqrt{x(x^2 +1)}}$
Dangerous one.
See:
http://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt%28x%28x%5E2%2B1%29%29
I've 2 different solutions:
1) This
http://integrals.wolfram.com/index.jsp?expr=1%2F(x*Sqrt[x^2+%2B1])
2) And this (provided by my professor)
$\displaystyle -\ln (x^-1 + \sqrt{x^-2 + 1})$
Wolfram says otherwise:
http://www.wolframalpha.com/input/?i=differentiate+-ln%28%281%2Fx%29+%2B+sqrt%281+%2B+x^%28-2%29%29%29
ie. $\displaystyle \frac{d}{dx} \left(-\ln \left(x^{-1} + \sqrt{x^{-2} + 1}\right) \right) = \frac{1}{x \sqrt{1 + x^2}}$.. Which is close, but not the same :P
Probably an elementary primitive of the function doesn't exist, so that we can try to obtain at least a series expansion of it. Let's start with...
$\displaystyle (1+x^{2})^{-\frac{1}{2}} = \sum_{n=0}^{\infty} \binom {-\frac{1}{2}}{n} x^{2n}$ (1)
... where $\displaystyle |x|<1$ and...
$\displaystyle \binom {-\frac{1}{2}}{n} = \frac{(-\frac{1}{2})\cdot (-\frac{3}{2})\dots (-n + \frac{1}{2})}{n!}$ (2)
... and, performing integration, we obtain...
$\displaystyle \int \frac{dx}{\sqrt{x}\cdot \sqrt{1+x^{2}}}= \sqrt{x}\cdot \sum_{n=0}^{\infty} \binom {-\frac{1}{2}}{n} \frac{x^{2n}}{2n+\frac{1}{2}} + c$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$