We are asked to find 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 find 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, we compute the derivative of $u$ with respect to $x$:

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

We notice that $x^2 + 72 = x^2 + 81 - 9$, and we seek a form $\frac{A(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2} + \frac{B}{x \sin x + 9 \cos x}$, for some constant $A$. Let's consider the derivative of $\frac{x \cos x - 9 \sin x}{x \sin x + 9 \cos x}$.

$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$.

Let $f(x) = -x \cos x + 9 \sin x$ and $g(x) = x \sin x + 9 \cos x$. Then

$f'(x) = -\cos x + x \sin x + 9 \cos x = x \sin x + 8 \cos x$

$g'(x) = \sin x + x \cos x - 9 \sin x = x \cos x - 8 \sin x$.

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

The numerator is

$x^2 \sin^2 x + 9x \sin x \cos x + 8x \sin x \cos x + 72 \cos^2 x - (-x^2 \cos^2 x + 8x \cos x \sin x + 9x \sin x \cos x - 72 \sin^2 x) =$

$x^2 \sin^2 x + 17x \sin x \cos x + 72 \cos^2 x + x^2 \cos^2 x - 17x \sin x \cos x + 72 \sin^2 x =$

$x^2(\sin^2 x + \cos^2 x) + 72(\cos^2 x + \sin^2 x) = x^2 + 72$.

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

3. Final Answer

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