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

Options

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

Answer

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

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

### Step 2: Expand the Square
Next, we expand the square in the integrand:
\[ I = \int \frac{(\ln x - 1)^2}{(1 + (\ln x)^2)^2} \, dx \]
This can be simplified to:
\[ I = \int \frac{\ln^2 x - 2\ln x + 1}{(1 + (\ln x)^2)^2} \, dx \]

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

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

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

### Step 6: Recognize the Derivative
Notice that the derivative of \( \frac{1}{1 + t^2} \) is \( -\frac{2t}{(1 + t^2)^2} \). Thus, we can use integration by parts or recognize that:
\[ \int e^t f(t) \, dt = e^t f(t) + C \]
where \( f(t) \) is a function whose derivative appears in the integrand.

### Step 7: Final Result
After evaluating the integrals and substituting back \( t = \ln x \), we get:
\[ I = \frac{x}{1 + (\ln x)^2} + C \]

### Conclusion
Thus, the final result of the integral is:
\[ \int \left( \frac{\ln x - 1}{1 + (\ln x)^2} \right)^2 \, dx = \frac{x}{1 + (\ln x)^2} + C \]