When facing integrals with inverse trigonometric functions, substituting the argument of the inverse function often helps simplify the expression. Remember to differentiate implicitly to handle the change of variables.
We begin by letting the given integral be:
\[
I = \int \frac{\sin^{-1} \left( \frac{x}{\sqrt{a + x}} \right)}{ \sqrt{a + x}} \, dx
\]
To simplify this, let us use the substitution:
\[
u = \sin^{-1} \left( \frac{x}{\sqrt{a + x}} \right)
\]
Then:
\[
\sin(u) = \frac{x}{\sqrt{a + x}}
\]
Square both sides to eliminate the square root:
\[
\sin^2(u) = \frac{x^2}{a + x}
\]
Now differentiate both sides with respect to \( x \):
\[
2 \sin(u) \cos(u) \frac{du}{dx} = \frac{2x}{a + x} - \frac{x^2}{(a + x)^2}
\]
Now, express the integrand using this substitution and simplify further to solve the integral.
