Summation: $x+ \sum _{ n=1 }^{ \infty }{ \frac { (-1)^{n-1} (2n-3) (2n-5)\cdots 5\cdot 3\cdot 1}{ 2^{n} n! (2n+1)}x^{2n+1} }$ I have to calculate:
$$x+ \sum _{ n=1 }^{ \infty }{ \frac { (-1)^{n-1} (2n-3) (2n-5)\cdots 5\cdot 3\cdot 1}{ 2^{n} n! (2n+1)}x^{2n+1} } $$
We have:
$$(2n-3)!! = \frac{(2n-2)!!}{(2n-2)!!}(2n-3)!! = \frac{(2n-2)!}{(2n-2)!!} = \frac{(2n-2)!}{2^{n-1}(n-1)!}$$
So, the expression in the series is:
$$\sum_{n\geq 1} (-1)^{n-1} \frac{(2n-2)!}{n!(n-1)! 2^{2n-1}(2n+1)}x^{2n+1} = -2x\sum_{n\geq 0} \binom{2n}{n}\frac{1}{(n+1)(2n+3)}\left(-\frac{x^2}{4}\right)^{n+1}$$
By the ratio test, the series converges when $x^2 < 1$. 
We now evaluate the series:
$$S(v) = \sum_{n\geq 0}v^{n+1} \binom{2n}{n} \frac{1}{n+1} = \int \sum_{n\geq 0} v^n \binom{2n}{n}~dv=\int \frac{1}{\sqrt{1-4v}}~dv=- \frac12\sqrt{1-4v}$$
Letting $v = -x^2/4$ afterwards will give us an expression for the series $S(u)$. 
But now I am stuck.

Instead of using $v^{n+1}$, it would have been best to use $v^{2n+1}$. You almost got it; here is a complete solution:
The following is a well-know Taylor series:
$$\sum_{n=0}^\infty \binom{2n}{n}x^{2n+1}=\frac{x}{\sqrt{1-4x^2}}$$
Integrating both sides of this equation yields the equation
$$\sum_{n=0}^\infty \binom{2n}{n}\frac{x^{2n+2}}{2n+2}=\frac{1-\sqrt{1-4x^2}}{4}$$
and integrating both sides of this equation gives us
$$\sum_{n=0}^\infty \binom{2n}{n}\frac{x^{2n+3}}{(2n+2)(2n+3)}=\frac{4x-2x\sqrt{1-4x^2}-\arcsin(2x)}{16}$$
Multiplying both sides by $2$ and making an appropriate substitution with $x$ will give you the desired result.
