To solve the integral
\[
I = \int \frac{x + (\cos^{-1}(3x))^2}{\sqrt{1 - 9x^2}} \, dx,
\]
we can break it down into two separate integrals:
\[
I = \int \frac{x}{\sqrt{1 - 9x^2}} \, dx + \int \frac{(\cos^{-1}(3x))^2}{\sqrt{1 - 9x^2}} \, dx.
\]
### Step 1: Solve the first integral
For the first integral, we will use the substitution:
\[
t = 1 - 9x^2 \implies dt = -18x \, dx \implies x \, dx = -\frac{dt}{18}.
\]
Now, we can rewrite the integral:
\[
\int \frac{x}{\sqrt{1 - 9x^2}} \, dx = \int \frac{x}{\sqrt{t}} \left(-\frac{dt}{18}\right) = -\frac{1}{18} \int \frac{1}{\sqrt{t}} \, dt.
\]
The integral of \( \frac{1}{\sqrt{t}} \) is \( 2\sqrt{t} \):
\[
-\frac{1}{18} \cdot 2\sqrt{t} = -\frac{1}{9} \sqrt{t} = -\frac{1}{9} \sqrt{1 - 9x^2}.
\]
### Step 2: Solve the second integral
For the second integral, we use the substitution:
\[
u = \cos^{-1}(3x) \implies x = \frac{\cos(u)}{3} \implies dx = -\frac{\sin(u)}{3} \, du.
\]
Now, we can rewrite the integral:
\[
\int \frac{(\cos^{-1}(3x))^2}{\sqrt{1 - 9x^2}} \, dx = \int \frac{u^2}{\sqrt{1 - 9\left(\frac{\cos(u)}{3}\right)^2}} \left(-\frac{\sin(u)}{3}\right) \, du.
\]
Calculating \( 1 - 9\left(\frac{\cos(u)}{3}\right)^2 \):
\[
1 - 9\left(\frac{\cos(u)}{3}\right)^2 = 1 - \cos^2(u) = \sin^2(u).
\]
Thus, \( \sqrt{1 - 9x^2} = \sin(u) \). The integral becomes:
\[
-\frac{1}{3} \int u^2 \, du.
\]
The integral of \( u^2 \) is \( \frac{u^3}{3} \):
\[
-\frac{1}{3} \cdot \frac{u^3}{3} = -\frac{1}{9} u^3.
\]
### Step 3: Combine the results
Now, combining both parts, we have:
\[
I = -\frac{1}{9} \sqrt{1 - 9x^2} - \frac{1}{9} (\cos^{-1}(3x))^3 + C.
\]
### Step 4: Compare with the given form
The solution can be expressed as:
\[
I = P\sqrt{1 - 9x^2} + Q(\cos^{-1}(3x))^3 + C,
\]
where \( P = -\frac{1}{9} \) and \( Q = -\frac{1}{9} \).
### Final Answer
Thus, the values of \( P \) and \( Q \) are:
\[
P = -\frac{1}{9}, \quad Q = -\frac{1}{9}.
\]
