Question

Find: \[ \int \frac{1}{x \left[(\log x)^2 - 3 \log x - 4\right]} \, dx. \]

Hint:

For integrals with quadratic denominators, use partial fractions after factorizing.

Answer 1

Step 1: Simplify the quadratic in the denominator.
Let \( u = \log x \). Then: \[ du = \frac{1}{x} \, dx. \] The integral becomes: \[ \int \frac{1}{x \left[(\log x)^2 - 3 \log x - 4\right]} \, dx = \int \frac{1}{(u^2 - 3u - 4)} \, du. \] Step 2: Factorize the quadratic.
Factorize \( u^2 - 3u - 4 \): \[ u^2 - 3u - 4 = (u - 4)(u + 1). \] The integral becomes: \[ \int \frac{1}{(u - 4)(u + 1)} \, du. \] Step 3: Use partial fraction decomposition.
Express \( \frac{1}{(u - 4)(u + 1)} \) as: \[ \frac{1}{(u - 4)(u + 1)} = \frac{A}{u - 4} + \frac{B}{u + 1}. \] Solve for \( A \) and \( B \): \[ 1 = A(u + 1) + B(u - 4). \] Let \( u = 4 \): \[ 1 = A(4 + 1) \quad \Rightarrow \quad A = \frac{1}{5}. \] Let \( u = -1 \): \[ 1 = B(-1 - 4) \quad \Rightarrow \quad B = -\frac{1}{5}. \] Thus: \[ \frac{1}{(u - 4)(u + 1)} = \frac{\frac{1}{5}}{u - 4} - \frac{\frac{1}{5}}{u + 1}. \] Step 4: Integrate each term.
The integral becomes: \[ \int \frac{1}{(u - 4)(u + 1)} \, du = \frac{1}{5} \int \frac{1}{u - 4} \, du - \frac{1}{5} \int \frac{1}{u + 1} \, du. \] Evaluate: \[ \int \frac{1}{u - 4} \, du = \ln|u - 4|, \quad \int \frac{1}{u + 1} \, du = \ln|u + 1|. \] Thus: \[ \int \frac{1}{(u - 4)(u + 1)} \, du = \frac{1}{5} \ln|u - 4| - \frac{1}{5} \ln|u + 1| + C. \] Step 5: Back-substitute \( u = \log x \).
Substitute \( u = \log x \) back into the result: \[ \int \frac{1}{x \left[(\log x)^2 - 3 \log x - 4\right]} \, dx = \frac{1}{5} \ln|\log x - 4| - \frac{1}{5} \ln|\log x + 1| + C. \] Conclusion:
The value of the integral is: \[ \boxed{\frac{1}{5} \ln|\log x - 4| - \frac{1}{5} \ln|\log x + 1| + C}. \]