Question:

If \[ \int \frac{1}{x(\log x)^2 + 4\log x - 1} \, dx = A \log [ \log x + B ] + K \] where \( K \) is the constant of integration, then \[ \int \frac{1}{x(\log x)^2 + 4\log x - 1} \, dx = A \log [ \log x + C ] + K. \]

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When dealing with logarithmic integrals, consider substitution to simplify the expression and solve for the constants step by step.
Updated On: May 21, 2025
  • \( A = \frac{1}{2\sqrt{5}}, B = (2 - \sqrt{5}), C = (2 + \sqrt{5}) \)
  • \( A = -\frac{1}{2\sqrt{5}}, B = (2 - \sqrt{5}), C = (2 + \sqrt{5}) \)
  • \( A = \frac{1}{2\sqrt{5}}, B = (2 + \sqrt{5}), C = (2 - \sqrt{5}) \)
  • \( A = -\frac{1}{2\sqrt{5}}, B = (2 + \sqrt{5}), C = (2 - \sqrt{5}) \)
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The Correct Option is A

Approach Solution - 1

Given the integral and its result as \( A \log [ \log x + B ] + K \), we need to solve for \( A \), \( B \), and \( C \). Step 1: The structure of the integral suggests a substitution involving the logarithmic terms. Step 2: By simplifying and equating terms, we find the values of \( A \), \( B \), and \( C \) to be: \[ A = \frac{1}{2\sqrt{5}}, \quad B = 2 - \sqrt{5}, \quad C = 2 + \sqrt{5}. \] % Final Answer The correct values are \( A = \frac{1}{2\sqrt{5}}, B = (2 - \sqrt{5}), C = (2 + \sqrt{5}) \).
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Approach Solution -2

Step 1: Simplify the integral expression
Let \( t = \log x \). Then \( dt = \frac{1}{x} dx \) or \( dx = x \, dt \).

Step 2: Rewrite the integral in terms of \( t \)
\[ \int \frac{1}{x (t^2 + 4t - 1)} \, dx = \int \frac{1}{t^2 + 4t - 1} \, dt \]

Step 3: Factor or complete the square for the denominator
\[ t^2 + 4t - 1 = (t^2 + 4t + 4) - 5 = (t + 2)^2 - 5 \]

Step 4: Use partial fraction or substitution
Rewrite the integral:
\[ \int \frac{1}{(t+2)^2 - (\sqrt{5})^2} dt = \int \frac{1}{(t+2 - \sqrt{5})(t+2 + \sqrt{5})} dt \]

Step 5: Use partial fractions
Set \[ \frac{1}{(t+2 - \sqrt{5})(t+2 + \sqrt{5})} = \frac{A}{t+2 - \sqrt{5}} + \frac{B}{t+2 + \sqrt{5}} \]
Multiply both sides by denominator:
\[ 1 = A(t+2 + \sqrt{5}) + B(t+2 - \sqrt{5}) \]

Step 6: Solve for A and B
Put \( t = -2 + \sqrt{5} \):
\[ 1 = A(2\sqrt{5}) \implies A = \frac{1}{2\sqrt{5}} \]
Put \( t = -2 - \sqrt{5} \):
\[ 1 = B(-2\sqrt{5}) \implies B = -\frac{1}{2\sqrt{5}} \]

Step 7: Write the integral
\[ \int \frac{1}{t^2 + 4t -1} dt = \frac{1}{2\sqrt{5}} \int \frac{1}{t + 2 - \sqrt{5}} dt - \frac{1}{2\sqrt{5}} \int \frac{1}{t + 2 + \sqrt{5}} dt \]

Step 8: Integrate
\[ = \frac{1}{2\sqrt{5}} \log | t + 2 - \sqrt{5} | - \frac{1}{2\sqrt{5}} \log | t + 2 + \sqrt{5} | + K \]

Step 9: Combine logarithms
\[ = \frac{1}{2\sqrt{5}} \log \left| \frac{t + 2 - \sqrt{5}}{t + 2 + \sqrt{5}} \right| + K \]

Step 10: Substitute back \( t = \log x \)
\[ \int \frac{1}{x(\log x)^2 + 4\log x - 1} dx = \frac{1}{2\sqrt{5}} \log \left| \frac{\log x + 2 - \sqrt{5}}{\log x + 2 + \sqrt{5}} \right| + K \]

Step 11: Express in the form \( A \log[\log x + B] + K \)
Notice the integral can be expressed as:
\[ A \log[\log x + B] + K \quad \text{with} \quad A = \frac{1}{2\sqrt{5}}, \quad B = 2 - \sqrt{5}, \quad C = 2 + \sqrt{5} \]

Final answer:
\[ A = \frac{1}{2\sqrt{5}}, \quad B = 2 - \sqrt{5}, \quad C = 2 + \sqrt{5} \]
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