Can you solve it for me?

We can write (using the substitution $x = -u$ and $x = \sin\theta$ ) $\begin{aligned} f(k) \; = \; \int_{-1}^1 \frac{\sqrt{1-x^2}}{\sqrt{k+1}-x}\,dx & = \; \int_{-1}^1 \frac{\sqrt{1-u^2}}{\sqrt{k+1}+u}\,du \\ & = \; \frac12\left(\int_{-1}^1 \frac{\sqrt{1-x^2}}{\sqrt{k+1}-x}\,dx + \int_{-1}^1 \frac{\sqrt{1-x^2}}{\sqrt{k+1}+x}\,dx\right) \; = \; \sqrt{k+1}\int_{-1}^1 \frac{\sqrt{1-x^2}}{k+1 - x^2}\,dx \\ & = \; \sqrt{k+1} \int_{-\frac12\pi}^{\frac12\pi} \frac{\cos^2\theta}{k + \cos^2\theta}\,d\theta \; = \; \pi\sqrt{k+1} - k\sqrt{k+1}\int_{-\frac12\pi}^{\frac12\pi} \frac{d\theta}{k + \cos^2\theta} \\ & = \; \pi\sqrt{k+1} - 2k\sqrt{k+1}\int_{-\frac12\pi}^{\frac12\pi} \frac{d\theta}{2k + 1 + \cos2\theta} \; = \; \pi\sqrt{k+1} - k\sqrt{k+1}\int_{-\pi}^{\pi} \frac{d\theta}{2k + 1 + \cos\theta} \\ & =\; \pi\sqrt{k+1} - \frac{2k\sqrt{k+1}}{i}\int_{|z|=1} \frac{dz}{z^2 + 2(2k+1)z + 1} \; =\; \pi\sqrt{k+1} - \frac{2k\sqrt{k+1}}{i} \times \frac{2\pi i}{4\sqrt{k(k+1)}} \\ & = \; \pi(\sqrt{k+1} - \sqrt{k}) \end{aligned}$ after some standard contour integration. Thus $S = \pi(\sqrt{100}-\sqrt{0}) = \boxed{10}\pi$ .