(a) $2 \sqrt{1-x^2}$,  $\frac{1}{2}\left(\sin ^{-1} x+x \sqrt{1-x^2}\right)+C$

(b) $\frac{1}{6} \sqrt{4 x^3+1}+C$

(a)

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1} x+x \sqrt{1-x^2}\right) &=\frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}+x\left(\frac{-2 x}{2 \sqrt{1-x^2}}\right) \\ &=\frac{1-x^2}{\sqrt{1-x^2}}+\sqrt{1-x^2} \\ &=\sqrt{1-x^2}+\sqrt{1-x^2} \\ &=2 \sqrt{1-x^2} \end{aligned}$

$\int \sqrt{1-x^2} \mathrm{~d} x=\frac{1}{2}\left(\sin ^{-1} x+x \sqrt{1-x^2}\right)+C$

(b) $\int \frac{x^2}{\sqrt{4 x^3+1}} \mathrm{~d} x =\frac{1}{12} \int 12 x^2\left(4 x^3+1\right)^{-\frac{1}{2}} \mathrm{~d} x
=\frac{1}{12} \frac{(4 x^3+1)^{\frac{1}{2}}}{\frac{1}{2}}+C
=\frac{1}{6} \sqrt{4 x^3+1}+C$