Hi

Evaluate:

$\displaystyle \int \frac{1}{\sqrt{x(x^2 +1)}}$

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- Mar 16th 2010, 11:44 AMShockquite difficult integral
Hi

Evaluate:

$\displaystyle \int \frac{1}{\sqrt{x(x^2 +1)}}$ - Mar 16th 2010, 12:26 PMGeneral
**Dangerous one.**

**See:**

**http://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt%28x%28x%5E2%2B1%29%29** - Mar 16th 2010, 12:49 PMShock
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)

$\displaystyle -\ln (x^-1 + \sqrt{x^-2 + 1})$ - Mar 16th 2010, 12:57 PMGeneral
**Use the substitution $\displaystyle x=tan(\theta)$ ..**

**to get (with some trig-identities):**

**$\displaystyle \sqrt{2} \int \frac{d\theta}{\sqrt{sin(2\theta})}$**

**Integration by parts maybe?** - Mar 16th 2010, 01:23 PMDefunkt
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 - Mar 16th 2010, 02:13 PMchisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$