To solve the integral
\[
I = \int \frac{x \tan x \sec x}{(\tan x - x)^2} \, dx,
\]
we will follow a step-by-step approach.
### Step 1: Rewrite the integrand
We start by rewriting the integrand in a more manageable form. Recall that
\[
\tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \sec x = \frac{1}{\cos x}.
\]
Thus, we can rewrite the integrand as:
\[
I = \int \frac{x \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}}{(\frac{\sin x}{\cos x} - x)^2} \, dx = \int \frac{x \sin x}{\cos^2 x \left(\frac{\sin x - x \cos x}{\cos x}\right)^2} \, dx.
\]
### Step 2: Simplify the denominator
Now, we simplify the denominator:
\[
\frac{\sin x - x \cos x}{\cos x} = \sin x - x \cos x.
\]
Thus, we can rewrite the integral as:
\[
I = \int \frac{x \sin x}{\cos^2 x (\sin x - x \cos x)^2} \, dx.
\]
### Step 3: Substitution
Next, we will use substitution. Let
\[
t = \sin x - x \cos x.
\]
Now, we differentiate \(t\) with respect to \(x\):
\[
dt = \left(\cos x - (-x \sin x + \cos x)\right) dx = (\cos x + x \sin x - \cos x) \, dx = x \sin x \, dx.
\]
Thus, we have:
\[
x \sin x \, dx = dt.
\]
### Step 4: Substitute in the integral
Now, substituting \(t\) into the integral gives us:
\[
I = \int \frac{dt}{t^2}.
\]
### Step 5: Integrate
The integral of \(t^{-2}\) is:
\[
\int t^{-2} \, dt = -\frac{1}{t} + C.
\]
### Step 6: Substitute back
Now we substitute back \(t\):
\[
I = -\frac{1}{\sin x - x \cos x} + C.
\]
### Step 7: Final answer
To express it in a more standard form, we can multiply the numerator and denominator by -1:
\[
I = \frac{1}{x \cos x - \sin x} + C.
\]
### Conclusion
Thus, the final answer is:
\[
\int \frac{x \tan x \sec x}{(\tan x - x)^2} \, dx = \frac{1}{x \cos x - \sin x} + C.
\]
