Seeking Methods to solve $F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$ I'm looking for different methods to solve the following integral. 
$$ F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$$
For $\alpha > 0$
Here the method I took was to employ integration by parts and then call to special functions, but can this equally be achieved with say a Feynman Trick? or another form integral transform?
My approach in detail:
Employ integration by parts:
\begin{align}
    v'(x) &= x^\alpha & u(x) &= \arcsin(x) \\
    v(x) &= \frac{x^{\alpha + 1}}{\alpha + 1} & u'(x) &= \frac{1}{\sqrt{1 - x^2}}
\end{align}
Thus,
\begin{align}
    F\left(\alpha\right) &= \left[\frac{x^{\alpha + 1}}{\alpha + 1}\cdot\arcsin(x)\right]_0^1 - \int_0^1 \frac{x^{\alpha + 1}}{\alpha + 1} \cdot \frac{1}{\sqrt{1 - x^2}} \:dx \\
    &= \frac{\pi}{2\left(\alpha + 1\right)} -  \frac{1}{\alpha + 1}\int_0^1 x^{\alpha + 1}\left(1 - x^2\right)^{-\frac{1}{2}} \:dx
\end{align}
Here make the substitution $u = x^2$ to obtain
\begin{align}
     F\left(\alpha\right) &= \frac{\pi}{2\left(\alpha + 1\right)} -  \frac{1}{\alpha + 1}\int_0^1 \left(\sqrt{u}\right)^{\alpha + 1}\left(1 - u\right)^{-\frac{1}{2}} \frac{\:du}{2\sqrt{u}} \\
     &= \frac{\pi}{2\left(\alpha + 1\right)} -  \frac{1}{2\left(\alpha + 1\right)}\int_0^1 u^{\frac{\alpha}{2}}\left(1 - u\right)   ^{-\frac{1}{2}} \:du     \\   
     &= \frac{1}{2\left(\alpha + 1\right)} \left[ \pi  - B\left(\frac{\alpha + 2}{2}, \frac{1}{2} \right) \right]
\end{align}
\begin{align}
    F\left(\alpha\right) &=\frac{1}{2\left(\alpha + 1\right)} \left[ \pi  - \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{\alpha + 2}{2} + \frac{1}{2}\right)} \right] \\
    &= \frac{1}{2\left(\alpha + 1\right)} \left[ \pi  - \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)\sqrt{\pi}}{\Gamma\left(\frac{\alpha + 3}{2}\right) } \right] \\
    &= \frac{\sqrt{\pi}}{2\left(\alpha + 1\right)} \left[ \sqrt{\pi}  - \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)}{\Gamma\left(\frac{\alpha + 3}{2}\right) } \right]
\end{align}
Edits:
Correction of original limit observation (now removed)
Correction of not stating region of convergence for $\alpha$.
Correction of 1/sqrt to sqrt in final line. 
Thanks to those commentators for pointing out. 
 A: Here is a method that relies on using a double integral.
Noting the integral converges for $\alpha > -2$, recognising
$$\arcsin x = \int_0^x \frac{du}{\sqrt{1 - u^2}},$$
the integral can be rewritten as
$$I = \int_0^1 \int_0^x \frac{x^\alpha}{\sqrt{1 - u^2}} \, du dx.$$
On changing the order of integration, one has
\begin{equation}
\int_0^1 \int_u^1 \frac{x^\alpha}{\sqrt{1 - u^2}} \, dx du. \qquad (*)
\end{equation}
After performing the $x$-integral we are left with
$$I = \frac{1}{\alpha + 1} \int_0^1 \frac{1 - u^{\alpha + 1}}{\sqrt{1 - u^2}} \, du, \quad \alpha \neq -1.$$
Enforcing a substitution of $u \mapsto \sqrt{u}$ results in
$$I = \frac{1}{2(\alpha + 1)} \int_0^1 \left (\frac{1}{\sqrt{u(1 - u)}} - \frac{u^{\alpha/2}}{\sqrt{ 1 - u}} \right ) \, du = I_1 - I_2.$$
The first of the integrals is trivial
$$I_1 = \frac{1}{2(\alpha + 1)} \int_0^1 \frac{du}{\sqrt{\frac{1}{4} - (u - \frac{1}{2})^2}} = \frac{1}{2(\alpha + 1)} \arcsin (2u - 1) \Big{|}^1_0 = \frac{\pi}{2(\alpha + 1)}.$$
For the second of the integrals, it can be evaluated by writing it in terms of the beta function. Here
\begin{align}
I_2 &= \int_0^1 \frac{u^{\alpha/2}}{\sqrt{1 - u}} \, du\\
&= \int_0^1 u^{(\alpha/2 + 1) - 1} (1 - u)^{1/2 - 1} \, du\\
&= \text{B} \left (\frac{\alpha}{2} + 1, \frac{1}{2} \right )\\
&= \sqrt{\pi} \, \frac{\Gamma \left (\frac{\alpha + 2}{2} \right )}{\Gamma \left (\frac{\alpha + 3}{2} \right )}.\\
\end{align}
Thus
$$I = \frac{1}{2(\alpha + 1)} \left [\pi - \sqrt{\pi} \, \frac{\Gamma \left (\frac{\alpha + 2}{2} \right )}{\Gamma \left (\frac{\alpha + 3}{2} \right )} \right ], \qquad \alpha \neq -1.$$
For the case when $\alpha = -1$, the double integral at ($*$) becomes
$$I = \int_0^1 \int_u^1 \frac{1}{x\sqrt{1 - u^2}} \, dx du.$$
After performing the $x$-integral which yields a natural logarithm, one has
$$I = -\int_0^1 \frac{\ln u}{\sqrt{1 - u^2}} \, du.$$
Enforcing a substitution of $u \mapsto \sin u$ leads to
$$I = -\int_0^{\pi/2} \ln (\sin u) \, du. \qquad (**)$$
The integral appearing in ($**$) is quite famous and reasonably (?) well known. Its evaluation can be found either here or here. Thus
$$I = \frac{\pi}{2} \ln 2, \qquad \alpha = -1.$$ 
A: Answer 2.0:
We know that for $|x|<1$, 
$$\arcsin x=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)}x^{2k+1}$$
Where $$(1/2)_k=\frac{\Gamma(1/2+k)}{\Gamma(1/2)}$$
Hence we may begin with 
$$F(a)=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)}\int_0^1x^{2k+a+1}\mathrm dx=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)(2k+2+a)}$$
Then we play with the fractions a little to get
$$F(a)=\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!}\bigg[\frac1{2k+1}-\frac1{2k+2+a}\bigg]$$
$$F(a)=\frac1{a+1}\sum_{n\geq0}\frac{(1/2)_n}{n!}\frac1{2n+1}-\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!}\frac1{2k+2+a}$$
$$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+2+a)}$$
$$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!}\int_0^1x^{2k+1+a}\mathrm dx$$
$$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\int_0^1\sum_{k\geq0}\frac{(1/2)_k}{k!}x^{2k+1+a}\mathrm dx$$
$$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\int_0^1\frac{x^{a+1}}{\Gamma(1/2)}\sum_{k\geq0}\frac{\Gamma(1/2+k)}{k!}x^{2k}\mathrm dx$$
$$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\int_0^1\frac{x^{a+1}}{\sqrt{1-x^2}}\mathrm dx$$
$u=x^2$:
$$F(a)=\frac\pi{2(a+1)}-\frac1{2(a+1)}\int_0^1u^{a/2}(1-u)^{-1/2}\mathrm du$$
$$F(a)=\frac\pi{2(a+1)}-\frac{\Gamma(\frac{a+2}2)\Gamma(\frac12)}{2(a+1)\Gamma(\frac{a+3}2)}$$
$$F(a)=\frac{\sqrt{\pi}}{2(a+1)}\bigg[\sqrt{\pi}-\frac{\Gamma(\frac{a+2}2)}{\Gamma(\frac{a+3}2)}\bigg]$$
