`int (sin^2x-cos^2x)/(sin x cos x) dx=`_____

Answer

To solve the integral

\[ I = \int \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} \, dx, \]

we can break it down into two separate integrals.

### Step 1: Rewrite the integral
We can express the integral as:

\[ I = \int \left( \frac{\sin^2 x}{\sin x \cos x} - \frac{\cos^2 x}{\sin x \cos x} \right) \, dx. \]

### Step 2: Simplify each term
This simplifies to:

\[ I = \int \left( \frac{\sin x}{\cos x} - \frac{\cos x}{\sin x} \right) \, dx. \]

### Step 3: Rewrite in terms of trigonometric functions
Recognizing the terms, we can rewrite the integral as:

\[ I = \int \tan x \, dx - \int \cot x \, dx. \]

### Step 4: Integrate each term
Now we can integrate each term separately:

1. The integral of \(\tan x\) is \(-\ln |\cos x| + C_1\).
2. The integral of \(\cot x\) is \(\ln |\sin x| + C_2\).

Thus, we have:

\[ I = -\ln |\cos x| - \ln |\sin x| + C. \]

### Step 5: Combine the logarithms
Using the properties of logarithms, we can combine the two logarithmic terms:

\[ I = -\ln |\cos x \sin x| + C. \]

### Final Answer
Therefore, the final result of the integral is:

\[ I = -\ln |\cos x \sin x| + C. \]

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