We are asked to evaluate the indefinite integral of the function $\frac{x^2 + 72}{(x \sin x + 9 \cos x)^2}$ with respect to $x$.

1. Problem Description

We are asked to evaluate the indefinite integral of the function $\frac{x^2 + 72}{(x \sin x + 9 \cos x)^2}$ with respect to $x$.

2. Solution Steps

Let $u = x \sin x + 9 \cos x$.

Then $\frac{du}{dx} = \sin x + x \cos x - 9 \sin x = x \cos x - 8 \sin x$.

We can rewrite the numerator as:

$x^2 + 72 = x^2 + 81 - 9 = (x^2 + 81) - 9$.

Let's consider the derivative of $\frac{\cos x}{x \sin x + 9 \cos x}$:

$\frac{d}{dx} \left( \frac{\cos x}{x \sin x + 9 \cos x} \right) = \frac{-\sin x (x \sin x + 9 \cos x) - \cos x (x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{-x \sin^2 x - 9 \sin x \cos x - x \cos^2 x + 8 \sin x \cos x}{(x \sin x + 9 \cos x)^2} = \frac{-x (\sin^2 x + \cos^2 x) - \sin x \cos x}{(x \sin x + 9 \cos x)^2} = \frac{-x - \sin x \cos x}{(x \sin x + 9 \cos x)^2}$. This does not appear useful.

Let us consider the derivative of $\frac{x}{x \sin x + 9 \cos x}$.

$\frac{d}{dx} \left( \frac{x}{x \sin x + 9 \cos x} \right) = \frac{1(x \sin x + 9 \cos x) - x(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{x \sin x + 9 \cos x - x^2 \cos x + 8x \sin x}{(x \sin x + 9 \cos x)^2} = \frac{9x \sin x + 9 \cos x - x^2 \cos x}{(x \sin x + 9 \cos x)^2}$.

Let $f(x) = \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2}$. We seek to find $F(x)$ such that $F'(x) = f(x)$.

Observe that $\frac{d}{dx} \left(\frac{\sin x - \frac{x}{9} \cos x}{x \sin x + 9 \cos x} \right) = \frac{(\cos x - \frac{1}{9} \cos x + \frac{x}{9} \sin x) (x \sin x + 9 \cos x) - (\sin x - \frac{x}{9} \cos x) (x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{(\frac{8}{9} \cos x + \frac{x}{9} \sin x) (x \sin x + 9 \cos x) - (\sin x - \frac{x}{9} \cos x) (x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} = \frac{\frac{8}{9} x \sin x \cos x + 8 \cos^2 x + \frac{x^2}{9} \sin^2 x + x \sin x \cos x - x \sin x \cos x + 8 \sin^2 x + \frac{x^2}{9} \cos^2 x - \frac{8}{9} x \sin x \cos x}{(x \sin x + 9 \cos x)^2} = \frac{8(\cos^2 x + \sin^2 x) + \frac{x^2}{9} (\sin^2 x + \cos^2 x)}{(x \sin x + 9 \cos x)^2} = \frac{8 + \frac{x^2}{9}}{(x \sin x + 9 \cos x)^2} = \frac{72+x^2}{9(x \sin x + 9 \cos x)^2}$.

Thus, $\int \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} dx = 9 \left( \frac{\sin x - \frac{x}{9} \cos x}{x \sin x + 9 \cos x} \right) + C = \frac{9 \sin x - x \cos x}{x \sin x + 9 \cos x} + C$.

3. Final Answer

$\frac{9 \sin x - x \cos x}{x \sin x + 9 \cos x} + C$