Integral of $\int{\frac{1}{\sqrt{x(1+x^2)}}dx}$ I was trying to solve the following question:

Evaluate: $$\int{\frac{1}{\sqrt{x(1+x^2)}}dx}$$

This is an unsolved question in my sample papers book and so I believe it should have an elementary primitive.
But, I don't know how to start this. I want a hint to get started with this. Thanks!

We can also express the solution in terms of the Beta Function and the Incomplete Beta Function as I cover here:
\begin{align}
I&= \int \frac{1}{\sqrt{x\left(1 + x^2\right)}}\:dx = \int_0^x \frac{t^{-\frac{1}{2}}}{\left(t^2 + 1 \right)^{\frac{1}{2}}} \:dt\\
&=\frac{1}{2} \left[ B\left(\frac{1}{2} - \frac{-\frac{1}{2} + 1}{2} , \frac{-\frac{1}{2} + 1}{2} \right) - B\left(\frac{1}{1 + x^2};\frac{1}{2} - \frac{-\frac{1}{2} + 1}{2} , \frac{-\frac{1}{2} + 1}{2} \right) \right] \\
&=\frac{1}{2} \left[ B\left( \frac{1}{4} , \frac{1}{4} \right) - B\left(\frac{1}{1 + x^2};\frac{1}{4} , \frac{1}{4}  \right) \right] 
\end{align}
Using the relationship between the Beta function and the Gamma Function we find that:
\begin{equation}
 B\left( \frac{1}{4} , \frac{1}{4} \right) = \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4} + \frac{1}{4}\right)} = \frac{\Gamma\left(\frac{1}{4}\right)^2}{\Gamma\left(\frac{1}{2}\right)} = \frac{\Gamma\left(\frac{1}{4}\right)^2}{\sqrt{\pi}} 
\end{equation}
Thus our integral $I$ becomes:
\begin{equation}
 I = \int \frac{1}{\sqrt{x\left(1 + x^2\right)}}\:dx  = \int_0^x \frac{t^{-\frac{1}{2}}}{\left(t^2 + 1 \right)^{\frac{1}{2}}} \:dt = \frac{1}{2} \left[ \frac{\Gamma\left(\frac{1}{4}\right)^2}{\sqrt{\pi}}  - B\left(\frac{1}{1 + x^2};\frac{1}{4} , \frac{1}{4}  \right) \right] 
\end{equation}
