how to integrate a potential function? I'm not getting integrate of this please help me. I started learning it not too long ago. then I'm not able to solve this yet. but I really need it
$$f'(x)=\frac{1}{\sqrt{1-x^2}}$$
 A: Recall the formula for a circle:
$$x^2+y^2=1$$
Solving for the upper half of the circle in terms of $x$,
$$y=\sqrt{1-x^2}$$
Now recall the formula for arc-length:
$$\operatorname{arc length}_{a,b}(f)=\int_a^b\sqrt{1+[f'(x)]^2}~\mathrm dx$$
Thus, the arc-length of a circle is given by
$$\begin{align}\operatorname{arc length}_{a,b}(\text{circle})&=\int_a^b\sqrt{1+[y'(x)]^2}~\mathrm dx\\&=\int_a^b\sqrt{1+\frac{x^2}{1-x^2}}~\mathrm dx\\&=\int_a^b\sqrt{\frac{1-x^2+x^2}{1-x^2}}~\mathrm dx\\&=\int_a^b\sqrt{\frac1{1-x^2}}~\mathrm dx\\&=\int_a^b\frac1{\sqrt{1-x^2}}~\mathrm dx\end{align}$$
Thus, it happens to be the case that $f(x)$ is equivalent to the arc-length of a circle plus a constant of integration.  We may set $a=0$ for simplicity and letting $b$ be our variable:
$$f(x)=\operatorname{arc length}_{0,x}(\text{circle})+C$$
The following red arc represents what we want to find:

Via some trigonometry, we find the purple angle is $\theta_x=\frac\pi2-\arctan\left(\frac{\sqrt{1-x^2}}x\right)$, and since the circumference of the full circle is $2\pi$, we derive that
$$f(x)=\theta_x+C=\frac\pi2-\arctan\left(\frac{\sqrt{1-x^2}}x\right)+C$$

$$f(x)=C^\star-\arctan\left(\frac{\sqrt{1-x^2}}x\right)$$

which holds for $x\in[0,1]$ and is equivalent to
$$f(x)=C^\star+\arcsin(x)$$
