`int((lnx-1)/((lnx)^(2)+1))^2dx` is equal to

Options

A. `(x)/(x^(2)+1)+C`
B. `(lnx)/((lnx)^(2)+1)`
C. `(x)/((lnx)^(2)+1)+C`
D. `e^(x)((x)/(x^(2)+1))+C`

Answer

To solve the integral \( I = \int \left( \frac{\ln x - 1}{(\ln x)^2 + 1} \right)^2 dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ I = \int \left( \frac{\ln x - 1}{(\ln x)^2 + 1} \right)^2 dx \]

### Step 2: Expand the Numerator
We can expand the square in the numerator:
\[ I = \int \frac{(\ln x - 1)^2}{((\ln x)^2 + 1)^2} dx \]
Expanding \((\ln x - 1)^2\) gives:
\[ (\ln x - 1)^2 = (\ln x)^2 - 2 \ln x + 1 \]
Thus,
\[ I = \int \frac{(\ln x)^2 - 2 \ln x + 1}{((\ln x)^2 + 1)^2} dx \]

### Step 3: Separate the Integral
We can separate the integral into three parts:
\[ I = \int \frac{(\ln x)^2}{((\ln x)^2 + 1)^2} dx - 2 \int \frac{\ln x}{((\ln x)^2 + 1)^2} dx + \int \frac{1}{((\ln x)^2 + 1)^2} dx \]

### Step 4: Substitution
Let \( t = \ln x \). Then \( dx = e^t dt \) and \( \ln x = t \). The integral becomes:
\[ I = \int \frac{t^2 e^t}{(t^2 + 1)^2} dt - 2 \int \frac{t e^t}{(t^2 + 1)^2} dt + \int \frac{e^t}{(t^2 + 1)^2} dt \]

### Step 5: Factor Out \( e^t \)
Now, we can factor out \( e^t \):
\[ I = e^t \left( \int \frac{t^2}{(t^2 + 1)^2} dt - 2 \int \frac{t}{(t^2 + 1)^2} dt + \int \frac{1}{(t^2 + 1)^2} dt \right) \]

### Step 6: Recognize the Form
Notice that the integrals can be recognized as derivatives:
Let \( f(t) = \frac{1}{1 + t^2} \). Then:
\[ f'(t) = -\frac{2t}{(1 + t^2)^2} \]
This gives us a relationship between the integrals.

### Step 7: Apply Integration by Parts
Using integration by parts or recognizing the pattern, we can integrate:
\[ \int e^t f(t) dt = e^t f(t) + C \]

### Step 8: Substitute Back
Substituting back \( t = \ln x \):
\[ I = e^{\ln x} \left( \frac{1}{1 + (\ln x)^2} \right) + C = \frac{x}{1 + (\ln x)^2} + C \]

### Final Answer
Thus, the solution to the integral is:
\[ I = \frac{x}{1 + (\ln x)^2} + C \]