$\text { The general solution of } \dfrac{d y}{d x}=\sin ^{-1} x \text { is }$

The given differential equation is $\dfrac{dy}{dx} = \sin^{-1} x$.

Integrating both sides with respect to $x$, we get $y = \int \sin^{-1} x \, dx$.

Using integration by parts, let $u = \sin^{-1} x$ and $dv = dx$. Then $du = \dfrac{1}{\sqrt{1-x^2}} dx$ and $v = x$.

The formula for integration by parts is $\int u \, dv = uv - \int v \, du$.

Substituting the values, we get $y = x \sin^{-1} x - \int \dfrac{x}{\sqrt{1-x^2}} dx$.

To evaluate the integral $\int \dfrac{x}{\sqrt{1-x^2}} dx$, let $t = 1 - x^2$. Then $dt = -2x \, dx$, which implies $x \, dx = -\dfrac{1}{2} dt$.

Substituting these into the integral, we get $\int \dfrac{x}{\sqrt{1-x^2}} dx = \int \dfrac{-1/2}{\sqrt{t}} dt = -\dfrac{1}{2} \int t^{-1/2} dt = -\dfrac{1}{2} \times \dfrac{t^{1/2}}{1/2} = -\sqrt{t} = -\sqrt{1-x^2}$.

Therefore, $y = x \sin^{-1} x - (-\sqrt{1-x^2}) + C$.

Simplifying, we get $y = x \sin^{-1} x + \sqrt{1-x^2} + C$.

Answer: $y=x \sin ^{-1} x+\sqrt{1-x^2}+C$