Is it possible to find elementary integral of $\frac{\arcsin x}{x}$? How could I possibly indefinite integral:
$$
\int{\arcsin\left(x\right) \over x}\,{\rm d}x
$$
using elementary functions? If it is impossible to integrate using elementary functions, Is there any way to find the definite integral in the range
$\left[0,\pi/2\right]$ ?. If so please teach me how to find it.
 A: As, Hurkyl suggests, substitute $x=\sin\theta$. I am  also assuming that you in fact intended the limits to be $0$ and $1$ since, $\arcsin$ is undefined for $\pi/2$. With some simple manipulations, the integral becomes
$$\int_0^{\pi/2}\frac{x\cos(x)}{\sin(x)}\,dx$$
Substitute, $t=\ln\sin(x)\implies dt=\cos(x)\,dx/\sin(x)$
$$\int_{-\infty}^0 x\,dt \text{ where $x = \arcsin(e^t)$}$$
Integrating by parts, i.e, applying $\int v\,du = uv - \int udv$
$$\lbrack x\ln(\sin x)\rbrack_0^{\pi/2} - \int_0^{\pi/2}\ln(\sin(x))dx$$
Both of these functions are elementary and solvable and left as an exercise to the reader. The answer is $\large \frac{\pi\ln(2)}{2}$

EDIT: I assumed that $\int_0^{\pi/2}\ln(\sin(x)dx$ would be familiar to most readers. I have since edited the answer to show the work here.
$$\int_0^{\pi/2}\ln(\sin(x))dx = \int_0^{\pi/2}\ln(\cos(x))dx$$
as noted in my comment. Adding the LHS to both sides, 
$$2\int_0^{\pi/2}\ln(\sin(x))dx = \int_0^{\pi/2}\ln\left(\frac{\sin(2x)}{2}\right)\,dx$$
$$2I = \int_0^{\pi/2}\ln\left(\sin(2x)\right) -\ln(2)\,dx$$
$$2I = \int_0^{\pi/2}\ln\left(\sin(2x)\right)\,dx -\pi\ln(2)/2$$
$$2I = \int_0^{\pi}\frac{\ln\left(\sin(u)\right)}{2}\,du -\pi\ln(2)/2$$
By symmetry arguments, $\int_0^{\pi}\ln(\sin(x))\,dx=2\int_0^{\pi/2}\ln(\sin(x))\,dx$
$$2I = \int_0^{\pi/2}\ln\left(\sin(u)\right)\,du -\pi\ln(2)/2$$
$$2I = I - \pi\ln(2)/2$$
$$I =  - \pi\ln(2)/2$$
A: You can find a antiderivative in terms of the dilogarithm.
I'm going to use the fact that $$\int_{0}^{z} f(x) \cot(x) \ dx = 2 \sum_{n=1}^{\infty} \int_{0}^{z} f(x) \sin(2nx) \ dx$$
and the fact that for $0 < x < 2 \pi,$
$$ \sum_{k=1}^{\infty} \frac{\cos kx}{k} = - \log \left( 2 \sin \frac{x}{2} \right)$$
Then
$$\int_{0}^{z} \frac{\arcsin x}{x} \ dx = \int_{0}^{\arcsin z} u \cot u \ du =$$
$$= 2 \sum_{n=1}^{\infty}\int_{0}^{\arcsin z} x \sin (2nx) \ dx = 2 \sum_{n=1}^{\infty} \Big(\frac{\sin (2nx)}{4n^{2}}  - \frac{z \cos (2nx)}{2n} \Big) \Big|^{\arcsin z}_{0}$$
$$ = \sum_{n=1}^{\infty}\frac{\sin(2n \arcsin z)}{2n^{2}} - \arcsin(z)\sum_{n=1}^{\infty}\frac{ \cos (2n \arcsin z)}{n}  $$
$$ =  \frac{1}{2} \text{Im} \ \text{Li}_{2} \left(e^{2i \arcsin z} \right) + \arcsin(z) \log(2z)$$
A: $\newcommand{\+}{^{\dagger}}
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With $\ds{\Lambda \in {\mathbb R}}$:
\begin{align}
\int_{0}^{\Lambda}{\arcsin\pars{x} \over x}\,\dd x&=
\sgn\pars{\Lambda}\int_{0}^{\verts{\Lambda}}{\arcsin\pars{x} \over x}\,\dd x
\\[3mm]&=
\arcsin\pars{\Lambda}\ln\pars{\verts{\Lambda}} -
\sgn\pars{\Lambda}\color{#f00}{\int_{0}^{\verts{\Lambda}}\ln\pars{x}\,{1 \over \root{1 - x^{2}}}\,\dd x}\tag{1}
\end{align}

With $\ds{x \equiv t^{1/2}\quad\imp\quad t = x^{2}}$
  \begin{align}
&\color{#f00}{\int_{0}^{\verts{\Lambda}}\ln\pars{x}\,{1 \over \root{1 - x^{2}}}
\,\dd x}=
\int_{0}^{\Lambda^{2}}\ln\pars{t^{1/2}}\,
{1 \over \root{1 - \pars{t^{1/2}}^{2}}}\,\half\,t^{-1/2}\dd t
\\[3mm]&=
{1 \over 4}\lim_{\mu \to -1/2}\totald{}{\mu}\int_{0}^{\Lambda^{2}}t^{\mu}
\pars{1 - t}^{-1/2}\,\dd t
={1 \over 4}\lim_{\mu \to -1/2}\totald{{\rm B}_{\Lambda^{2}}\pars{\mu + 1,1/2}}{\mu}
\end{align}
  where $\ds{{\rm B}_{z}\pars{a,b}}$ is the
  Incomplete Beta Function.
  By replacing in $\pars{1}$:
  $$
\color{#00f}{\int_{0}^{\Lambda}{\arcsin\pars{x} \over x}\,\dd x
=\arcsin\pars{\Lambda}\ln\pars{\verts{\Lambda}} - {1 \over 4}\,\sgn\pars{\Lambda}
\lim_{\mu \to -1/2}\totald{{\rm B}_{\Lambda^{2}}\pars{\mu + 1,1/2}}{\mu}}
$$

For the particular cases $\ds{\Lambda = \pm 1}$,
$\ds{{\rm B}_{\Lambda^{2}}\pars{\mu + 1,1/2} = {\rm B}_{1}\pars{\mu + 1,1/2}
={\rm B}\pars{\mu + 1,1/2}}$ where $\ds{{\rm B}\pars{a,b}}$ is the
Beta Function. Also,
$\ds{{\rm B}\pars{a,b} = \Gamma\pars{a}\Gamma\pars{b}/\Gamma\pars{a + b}}$.
$\ds{\Gamma\pars{z}}$ is the
Gamma Function.
Then,
\begin{align}
\int_{0}^{\pm 1}{\arcsin\pars{x} \over x}\,\dd x
&=\mp\,{1 \over 4}\lim_{\mu \to -1/2}\totald{}{\mu}\bracks{%
\Gamma\pars{\mu + 1}\Gamma\pars{1/2} \over \Gamma\pars{\mu + 3/2}}
\\[3mm]&=\mp\,{1 \over 4}\,\Gamma\pars{\half}\bracks{%
{\Gamma\pars{1/2}\Psi\pars{1/2} \over \Gamma\pars{1}}
-{\Gamma\pars{1/2}\Psi\pars{1} \over \Gamma\pars{1}}}
\\[3mm]&=\mp\,{1 \over 4}\,\Gamma^{2}\pars{\half}\bracks{\Psi\pars{\half} - \Psi\pars{1}}
\end{align}
$\ds{\Gamma\pars{1/2} = \root{\pi}}$. $\ds{\Psi\pars{1/2} = -\gamma - 2\ln\pars{2}.\ \Psi\pars{1} = -\gamma}$.
$\ds{\Psi\pars{z}}$ and $\ds{\gamma}$ are the Digamma Function and 
the Euler-Mascheroni Constant, respectively. See
this page. Then,
$$
\color{#00f}{\large\int_{0}^{\pm 1}{\arcsin\pars{x} \over x}\,\dd x
=
\pm\,\half\,\pi\ln\pars{2}}
$$
A: In a comment, I suggested to develop the integrand as an infinite Taylor expansion to be integrated later. Doing so, I obtained as a result 
$$\int \frac{\arcsin x}{x}\,dx=\sum _{n=0}^{\infty } \frac{4^{-n} (2 n)!}{(2 n+1)^2 (n!)^2} x^{2 n+1}$$ which has been later found to be the development of an hypergeometric function. So, as a final result, $$\int \frac{\arcsin x}{x}\,dx=x \,
   _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};x^2\right)$$ For sure, this result is not expressed on the basis of elementary functions but it is analytical.  
If the hypergeometric function cannot be used for any reason, the Taylor expansion can be efficiently used since it converges very quickly. As an example $$\int_{0}^{1} \frac{\arcsin x}{x} \ dx$$ is computed iwth a relative error of less than $0.1$% adding $20$ terms and this is probably more than sufficient for most engineering calculations.  
What is interesting is that this approach could be used for the calculation of
$$\int \frac{\arcsin x^a}{x^b}\,dx $$ where $a$ and $b$ could be any numbers.
