`int (x^(2)(1-"ln"x))/(("ln"^(4)x-x^(4))dx` is equal to

Answer

To solve the integral

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

we will follow these steps:

### Step 1: Simplify the Denominator
We can rewrite the denominator \(\ln^4 x - x^4\) by factoring out \(x^4\):

\[ \ln^4 x - x^4 = x^4 \left(\frac{\ln^4 x}{x^4} - 1\right). \]

### Step 2: Rewrite the Integral
Now, substituting this back into the integral, we get:

\[ I = \int \frac{x^2(1 - \ln x)}{x^4 \left(\frac{\ln^4 x}{x^4} - 1\right)} \, dx = \int \frac{(1 - \ln x)}{x^2 \left(\frac{\ln^4 x}{x^4} - 1\right)} \, dx. \]

### Step 3: Substitute \(t = \frac{\ln x}{x}\)
Let’s make the substitution \(t = \frac{\ln x}{x}\). Then, we differentiate:

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

### Step 4: Substitute in the Integral
Now, substituting \(t\) and \(dx\) into the integral:

\[ I = \int \frac{1}{t^4 - 1} dt. \]

### Step 5: Factor the Denominator
We can factor \(t^4 - 1\) as:

\[ t^4 - 1 = (t^2 - 1)(t^2 + 1) = (t - 1)(t + 1)(t^2 + 1). \]

### Step 6: Partial Fraction Decomposition
Now we can perform partial fraction decomposition:

\[ \frac{1}{(t - 1)(t + 1)(t^2 + 1)} = \frac{A}{t - 1} + \frac{B}{t + 1} + \frac{Ct + D}{t^2 + 1}. \]

### Step 7: Solve for Coefficients
Multiplying through by the denominator and equating coefficients will yield values for \(A\), \(B\), \(C\), and \(D\).

### Step 8: Integrate Each Term
Now we can integrate each term separately. The integrals will be:

- For \(\frac{A}{t - 1}\), it will yield \(A \ln |t - 1|\).
- For \(\frac{B}{t + 1}\), it will yield \(B \ln |t + 1|\).
- For \(\frac{Ct + D}{t^2 + 1}\), it will yield \(C \tan^{-1}(t) + D \ln(t^2 + 1)\).

### Step 9: Substitute Back
Finally, substitute back \(t = \frac{\ln x}{x}\) into the integrated result to express everything in terms of \(x\).

### Final Result
After performing all the integrations and substitutions, we will arrive at the final expression for the integral \(I\).

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