Question:

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

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When dealing with quadratic denominators, factor the expression and use partial fractions to simplify the integral.
Updated On: Jan 28, 2025
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Solution and Explanation

Step 1: Simplify the integral using substitution.
Let \( u = \log x \). Then, the derivative of \( u \) with respect to \( x \) is: \[ du = \frac{1}{x} \, dx. \] Substituting this into the integral, we get: \[ \int \frac{1}{x \left[(\log x)^2 - 3 \log x - 4\right]} \, dx = \int \frac{1}{(u^2 - 3u - 4)} \, du. \] Step 2: Factor the quadratic expression.
Factorize the quadratic expression \( u^2 - 3u - 4 \): \[ u^2 - 3u - 4 = (u - 4)(u + 1). \] Thus, the integral becomes: \[ \int \frac{1}{(u - 4)(u + 1)} \, du. \] Step 3: Perform partial fraction decomposition.
We decompose \( \frac{1}{(u - 4)(u + 1)} \) into partial fractions: \[ \frac{1}{(u - 4)(u + 1)} = \frac{A}{u - 4} + \frac{B}{u + 1}. \] Multiplying both sides by \( (u - 4)(u + 1) \) gives: \[ 1 = A(u + 1) + B(u - 4). \] Now, solve for \( A \) and \( B \): - 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, we have: \[ \frac{1}{(u - 4)(u + 1)} = \frac{\frac{1}{5}}{u - 4} - \frac{\frac{1}{5}}{u + 1}. \] Step 4: Integrate each term.
The integral now becomes: \[ \int \frac{1}{(u - 4)(u + 1)} \, du = \frac{1}{5} \int \frac{1}{u - 4} \, du - \frac{1}{5} \int \frac{1}{u + 1} \, du. \] The integrals are straightforward: \[ \int \frac{1}{u - 4} \, du = \ln|u - 4|, \quad \int \frac{1}{u + 1} \, du = \ln|u + 1|. \] Thus, the result is: \[ \int \frac{1}{(u - 4)(u + 1)} \, du = \frac{1}{5} \ln|u - 4| - \frac{1}{5} \ln|u + 1| + C. \] Step 5: Back-substitute \( u = \log x \).
Finally, substitute \( u = \log x \) back into the expression: \[ \int \frac{1}{x \left[(\log x)^2 - 3 \log x - 4\right]} \, dx = \frac{1}{5} \ln|\log x - 4| - \frac{1}{5} \ln|\log x + 1| + C. \] Conclusion:
Thus, the value of the integral is: \[ \boxed{\frac{1}{5} \ln|\log x - 4| - \frac{1}{5} \ln|\log x + 1| + C}. \] Using Log \(\frac{M}{N}\) = log M - log N
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