`int(x^(2)(1-logx))/((logx)^(4)-x^(4))dx` equals

Options

A. `(1)/(2)ln.(x)/(lnx)-(1)/(4)ln(ln^(2)x-x^(2))+C`
B. `(1)/(4)ln((lnx-x)/(lnx+x))-(1)/(2)tan^(-1)((lnx)/(x))+C`
C. `(1)/(4)ln((lnx-x)/(lnx+x))+(1)/(2)tan^(-1)((lnx)/(x))+C`
D. `(1)/(4)(ln((lnx-x)/(lnx+x))+tan^(-1)((lnx)/(x)))+C`

Explanation

To solve the integral

\[ I = \int \frac{x^2 (1 - \log x)}{\log^4 x - x^4} \, dx, \]

we will follow a series of steps to simplify and evaluate the integral.

### Step 1: Simplify the Denominator
We start by rewriting the denominator:

\[ \log^4 x - x^4 = (\log^2 x - x^2)(\log^2 x + x^2). \]

### Step 2: Rewrite the Integral
Now, we can rewrite the integral as:

\[ I = \int \frac{x^2 (1 - \log x)}{(\log^2 x - x^2)(\log^2 x + x^2)} \, dx. \]

### Step 3: Substitute \( t = \frac{\log x}{x} \)
To simplify the expression, we will use the substitution \( t = \frac{\log x}{x} \). Then, we differentiate:

\[ dt = \left( \frac{1}{x} - \frac{\log x}{x^2} \right) dx = \frac{1 - \log x}{x^2} \, dx. \]

This implies:

\[ dx = \frac{x^2}{1 - \log x} dt. \]

### Step 4: Substitute into the Integral
Substituting \( t \) and \( dx \) into the integral gives:

\[ I = \int \frac{x^2 (1 - \log x)}{(\log^2 x - x^2)(\log^2 x + x^2)} \cdot \frac{x^2}{1 - \log x} dt. \]

This simplifies to:

\[ I = \int \frac{x^4}{(\log^2 x - x^2)(\log^2 x + x^2)} dt. \]

### Step 5: Rewrite \( \log x \) in terms of \( t \)
Using the substitution \( t = \frac{\log x}{x} \), we can express \( \log x \) as \( x t \). Thus, we can rewrite our integral in terms of \( t \).

### Step 6: Factor the Denominator
Now, we can factor the denominator in terms of \( t \):

\[ \log^2 x = (xt)^2 = x^2 t^2. \]

This gives us:

\[ I = \int \frac{x^4}{(x^2 t^2 - x^2)(x^2 t^2 + x^2)} dt = \int \frac{x^2}{(t^2 - 1)(t^2 + 1)} dt. \]

### Step 7: Separate the Integral
We can separate the integral:

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

### Step 8: Integrate
Using standard results for integration:

1. \(\int \frac{dt}{t^2 - 1} = \frac{1}{2} \log \left| \frac{t - 1}{t + 1} \right| + C\)
2. \(\int \frac{dt}{t^2 + 1} = \tan^{-1}(t) + C\)

We can now evaluate the integral:

\[ I = \frac{1}{2} \left( \frac{1}{2} \log \left| \frac{t - 1}{t + 1} \right| - \tan^{-1}(t) \right) + C. \]

### Step 9: Substitute Back
Finally, substitute back \( t = \frac{\log x}{x} \):

\[ I = \frac{1}{4} \log \left| \frac{\frac{\log x}{x} - 1}{\frac{\log x}{x} + 1} \right| - \frac{1}{2} \tan^{-1}\left(\frac{\log x}{x}\right) + C. \]

### Final Result
Thus, the result of the integral is:

\[ I = \frac{1}{4} \log \left| \frac{\log x - x}{\log x + x} \right| - \frac{1}{2} \tan^{-1}\left(\frac{\log x}{x}\right) + C. \]