We need to evaluate the indefinite integral $\int \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} dx$.

1. Problem Description

We need to evaluate the indefinite integral $\int \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} dx$.

2. Solution Steps

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

Using the quotient rule, we have:

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

$= \frac{x \sin x \cos x + 9 \cos^2 x - \sin^2 x - x \sin x \cos x + 9 \sin^2 x}{(x \sin x + 9 \cos x)^2}$

$= \frac{9 (\cos^2 x + \sin^2 x) - (\sin^2 x)}{(x \sin x + 9 \cos x)^2} = \frac{9 - \sin^2 x}{(x \sin x + 9 \cos x)^2}$.

Now let's analyze the derivative of $\frac{-x \cos x}{x \sin x + 9 \cos x}$.

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

$= \frac{-x \sin x \cos x - 9 \cos^2 x + x^2 \sin^2 x + 9x \sin x \cos x + x \sin x \cos x + x^2 \cos^2 x - 9x \sin x \cos x}{(x \sin x + 9 \cos x)^2}$

$= \frac{-9 \cos^2 x + x^2 (\sin^2 x + \cos^2 x)}{(x \sin x + 9 \cos x)^2} = \frac{x^2 - 9 \cos^2 x}{(x \sin x + 9 \cos x)^2}$.

We notice that $\frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} = \frac{(x^2 - 9 \cos^2 x) + (9 - \sin^2 x) + 9 \cos^2 x + \sin^2 x + 63}{(x \sin x + 9 \cos x)^2} = \frac{x^2 - 9 \cos^2 x}{(x \sin x + 9 \cos x)^2} + \frac{9 - \sin^2 x}{(x \sin x + 9 \cos x)^2} + \frac{63}{(x \sin x + 9 \cos x)^2}$.

Let's consider $\frac{d}{dx} \left(\frac{\sin x - x \cos x}{x \sin x + 9 \cos x}\right)$.

Using the quotient rule, we have:

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

$= \frac{x^2 \sin^2 x + 9x \sin x \cos x - (\sin^2 x + x \sin x \cos x - 9 \sin^2 x - x \sin x \cos x - x^2 \cos^2 x + 9x \cos x \sin x)}{(x \sin x + 9 \cos x)^2}$

$= \frac{x^2 \sin^2 x + 9x \sin x \cos x - \sin^2 x - x \sin x \cos x + 9 \sin^2 x + x \sin x \cos x + x^2 \cos^2 x - 9x \sin x \cos x}{(x \sin x + 9 \cos x)^2}$

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

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

Hence we want to integrate $\int \frac{x^2 + 72}{(x \sin x + 9 \cos x)^2} dx$.

Let $u = \frac{-x \cos x + \sin x}{x \sin x + 9 \cos x}$.

Then $\frac{du}{dx} = \frac{x^2+8\sin^2(x)}{(x \sin(x)+9\cos(x))^2}$.

Consider $I = \int \frac{x^2+72}{(x \sin x + 9 \cos x)^2} dx = \int \frac{x^2+81 - 9+72}{(x \sin x + 9 \cos x)^2} dx$.

Let's try $\frac{x \sin x - 9 \cos x}{(x \sin x + 9 \cos x)}$.

Then $I = \frac{x \sin x - 9 \cos x}{(x \sin x + 9 \cos x)}$

Final guess:

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

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

Not quite.

$\frac{x \sin x -9 \cos x}{(x \sin x + 9 \cos x)}$.

3. Final Answer

$\frac{-x\cos(x)+\sin(x)}{x\sin(x)+9\cos(x)}$