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

1. Problem Description

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

2. Solution Steps

We will use integration by parts to solve the integral. Let's first rewrite the numerator as follows:

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

Then we can rewrite the integral as

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

Let $u = x \sin x + 9 \cos x$. Then the derivative of $u$ with respect to $x$ is:

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

Now, let's rewrite the integral in terms of $\sin x$ and $\cos x$. We can write the numerator as $x^2 + 72 = x^2 + 9^2 - 9$.

Let's try to manipulate the integral into the form $\int \frac{f'(x)}{[f(x)]^2} dx$. If $f(x) = x \sin x + 9 \cos x$, then $f'(x) = x \cos x - 8 \sin x$.

Now let us consider $\frac{Ax \cos x + B \sin x}{(x \sin x + 9 \cos x)}$. Its derivative is:

$\frac{(A \cos x - Ax \sin x + B \cos x)(x \sin x + 9 \cos x) - (Ax \cos x + B \sin x)(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2}$.

Let us assume the integral is of the form $\frac{Ax \cos x + B \sin x}{x \sin x + 9 \cos x}$. Then differentiating it, we get

$\frac{(x \sin x + 9 \cos x)(A \cos x - Ax \sin x + B \sin x + Bx \cos x) - (Ax \cos x + B \sin x)(x \cos x - 8 \sin x)}{(x \sin x + 9 \cos x)^2}$.

The numerator is

$(A \cos x - Ax \sin x + B \sin x + Bx \cos x)(x \sin x + 9 \cos x) - (Ax^2 \cos^2 x - 8Ax \sin x \cos x + Bx \sin x \cos x - 8B \sin^2 x)$

$Ax \sin x \cos x + 9A \cos^2 x - Ax^2 \sin^2 x - 9Ax \sin x \cos x + Bx \sin^2 x + 9B \sin x \cos x + Bx^2 \sin x \cos x + 9Bx \cos^2 x - Ax^2 \cos^2 x + 8Ax \sin x \cos x - Bx \sin x \cos x + 8B \sin^2 x$

Grouping the terms,

$Ax \sin x \cos x - 9Ax \sin x \cos x + 8Ax \sin x \cos x = 0$.

$9A \cos^2 x + Bx \sin^2 x + 9B \sin x \cos x + 9Bx \cos^2 x + 8B \sin^2 x - Ax^2 \sin^2 x - Ax^2 \cos^2 x$

$9A \cos^2 x + 9B \sin x \cos x + 8B \sin^2 x + Bx(\sin^2 x + x \cos^2 x) - Ax^2 (\sin^2 x + \cos^2 x)$

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

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

Numerator: $(-x \sin x \cos x + x^2 \sin^2 x -9x \sin x \cos x +9 \cos^2 x +x \sin x -81 \cos^2 x) - (-x^2 \cos^2 x +9x \sin x \cos x + 9x \sin x \cos x - 81 \sin^2 x)$

$= 9x \sin x \cos x + x^2 \sin^2 x + 81\cos^2 x = 9x \sin x \cos x - x^2 \cos^2 x -9x + 81 \sin^2 x$

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

3. Final Answer

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