What's the value of $\int \left(1-\sqrt{1-x^2} \right) dx$? $$\int \left(1-\sqrt{1-x^2} \right)  dx$$
Also, What is it's value from $0$ to $(1+\sqrt7)/2$ ?

I believe there might be missing pieces or a mistake in the inquiry, here's why:
let x = sin(u)
Thus, $\frac{dx}{du}=cos(u)$
$$\int 1-\sqrt{1-x^2}dx = \int (1-cos(u))(cos(u))du $$
Using the Double angle formula: $cos(2\theta)=2cos^2\theta -1$ , we have:
$$ sin(u) - \int\frac{cos(2u)-1}{2}du $$ which thus evaluates to (using double angle formula $sin(2\theta)=2sin(\theta)cos(\theta)$):
$$ sin(u)-\frac{1}{2}sin(u)cos(u)+\frac{1}{2}u + C $$
Substituting x back into the equation:
$$ x - \frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}sin^{-1}x+C $$
Coming back to the original integral:
$$ \int_{0}^{\frac{1+\sqrt{7}}{2}}1-\sqrt{1-x^2}dx = [x - \frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}sin^{-1}x]_{0}^{\frac{1+\sqrt{7}}{2}} $$
Here's where it doesn't make sense, the last part of the expression requires you to evaluate $$ sin^{-1}(\frac{1+\sqrt{7}}{2}) $$
which will not be possible since $\frac{1+\sqrt{7}}{2}$ lies outside the function domain of $sin^{-1}x$.
