Evaluate `int((log x-1)/(1+(logx)^(2)))^(2)dx`

Explanation

To evaluate the integral

\[ \int \left( \frac{\log x - 1}{1 + (\log x)^2} \right)^2 dx, \]

we will follow these steps:

### Step 1: Substitution
Let \( t = \log x \). Then, we have:

\[ x = e^t \quad \text{and} \quad dx = e^t dt. \]

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we get:

\[ \int \left( \frac{t - 1}{1 + t^2} \right)^2 e^t dt. \]

### Step 3: Expand the Numerator
Now, we will expand the numerator:

\[ \left( \frac{t - 1}{1 + t^2} \right)^2 = \frac{(t - 1)^2}{(1 + t^2)^2} = \frac{t^2 - 2t + 1}{(1 + t^2)^2}. \]

### Step 4: Substitute Back into the Integral
Thus, the integral becomes:

\[ \int \frac{(t^2 - 2t + 1)e^t}{(1 + t^2)^2} dt. \]

### Step 5: Separate the Integral
We can separate this into three integrals:

\[ \int \frac{t^2 e^t}{(1 + t^2)^2} dt - 2 \int \frac{t e^t}{(1 + t^2)^2} dt + \int \frac{e^t}{(1 + t^2)^2} dt. \]

### Step 6: Integration by Parts
For the second integral, we can use integration by parts. Let:

- \( u = \frac{1}{1 + t^2} \) and \( dv = e^t dt \).

Then, we have:

- \( du = -\frac{2t}{(1 + t^2)^2} dt \) and \( v = e^t \).

Applying integration by parts:

\[ \int u dv = uv - \int v du. \]

### Step 7: Combine Results
After performing the integration for each part, we will combine the results.

### Step 8: Substitute Back
Finally, we substitute \( t = \log x \) back into our result to express the answer in terms of \( x \).

### Final Answer
The final answer will be:

\[ \frac{x}{1 + (\log x)^2} + C, \]

where \( C \) is the constant of integration.

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