Another integral I need help with $\frac{1+\sqrt x}{\sqrt x\sqrt{1-x^2}} $ Can someone help me figure out how to compute the integral
$$\int \frac{1+\sqrt{x} } {\sqrt{x}\sqrt{1-x^2} }dx$$ 
Thanks in advance!
My attempts:
I tried substituting $\sqrt{x} = u $ , but it gives me nothing.
I also tried splitting the integral into the sum of two integrals, which leaves me with the calculation of the integral of $ \frac{1}{\sqrt{x}\sqrt{1-x^2} } $ , and I don't know how to calculate it.
 A: Your attempts failed with good reason.
Assuming everything's happening over $\mathbb{R}$, for this integral to make sense we must have $-1 \le x \le 1$ (in fact $0 \le x \le 1$), so a trigonometric substitution is valid. Putting $x = \sin \theta$, then we get
$$\begin{align}
\int \dfrac{1+\sqrt{x}}{\sqrt{x}\sqrt{1-x^2}}\, dx &= \int \dfrac{1+\sqrt{\sin \theta}}{\sqrt{\sin \theta} \cos \theta} \cdot \cos \theta\, d\theta \\
&= \int \dfrac{1+\sqrt{\sin \theta}}{\sqrt{\sin \theta}}\, d\theta \\
&= \int (\sin \theta)^{-\frac{1}{2}}\, d\theta + \theta + k\end{align}$$
for some constant of integration $k$.
But $(\sin \theta)^{-\frac{1}{2}}$ has no antiderivative expressible in terms of elementary functions. So there's no way of expressing the indefinite integral in terms of elementary functions.
A: HINT:
$$\int \frac{1+\sqrt{x} } {\sqrt{x}\sqrt{1-x^2} }dx$$ $\sqrt{x}=t$ then $\frac{dx}{\sqrt{x}}=2dt,x^2=t^4$
$$\int \frac{1+t } {\sqrt{1-t^4} }dt$$
A: It good to know that the expression of the form:
$$x^m(a+bx^n)^pdx$$
where $m,n,p,a,b$ are constant is called a differential binomial.

Theorem: The integral
$$\int x^m(a+bx^n)^pdx$$
can be reduced if $m,n,p$ are rational numbers, to the integral of a rational function, and can thus be expressed in terms of elementary functions if:
$1.$ $p$ is an integer( $p>0$ use the  Newton's binomial theorem and when $p<0$ then $x=t^k$ which $\text{lcm}(n,m)$).
$2.$ $\dfrac{m+1}{n}$ is an integer.
$3.$ $\dfrac{m+1}{n}+p$ is an integer.

It is clear that we should focus on the term $\int\frac{dx}{\sqrt{x}\sqrt{1-x^2}}$. By a simple investigation, above fact tells us this integral cannot be expressed in terms of elemantry functions. Try it!
A: The second part is
$$
\int\frac{\mathrm dx}{\sqrt{1-x^2}}\stackrel{(x=\sin\theta)}{=}\int\mathrm d\theta=\theta+C.
$$
The first part is
$$\int\frac{\mathrm dx}{\sqrt{x}\sqrt{1-x^2}}\stackrel{(x=\sqrt{y})}{=}\int\tfrac12y^{-3/4}(1-y)^{-1/2}\mathrm dy=\tfrac12\mathrm{B}(y;\tfrac14,\tfrac12)+C,
$$
where $\mathrm B$ denotes the incomplete Beta function.
Hence, on the interval $[0,1]$,
$$
\int\frac{1+\sqrt{x}}{\sqrt{x}\sqrt{1-x^2}}\mathrm dx=\tfrac12\mathrm{B}(x^2;\tfrac14,\tfrac12)+\arcsin(x)+C.
$$
For example,
$$
\int_0^1\frac{1+\sqrt{x}}{\sqrt{x}\sqrt{1-x^2}}\mathrm dx=\frac{\Gamma(\frac14)\Gamma(\frac12)}{2\Gamma(\frac34)}+\frac\pi2=\frac\pi2(\sqrt{2\pi}\Gamma(\tfrac14)^2+1)\approx1.7128.
$$
