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Class 12 Integrals · Long Answer3 Marksindefinite_integration

Class 12 Maths · Integrals · Long Answer #85

Evaluate \(\int \log{\left(\log{\left(x \right)} \right)} + \frac{1}{\log{\left(x \right)}^{2}}\,dx\). Work for x>1.

✅ Step-by-Step Solution

Final Answer: \(x \log{\left(\log{\left(x \right)} \right)} - \frac{x}{\log{\left(x \right)}}\) + C
  1. The real domain is x>1. Let u=ln x, so dx=e^u du.
  2. The integral becomes ∫e^u[ln u+u^(−2)]du. With g(u)=ln u−1/u, one has g+g′=ln u+u^(−2).
  3. Use d[e^u g(u)]/du=e^u(g+g′); substitute u=ln x to get xln(ln x)−x/ln x.
  4. Therefore the required integral is \(x \log{\left(\log{\left(x \right)} \right)} - \frac{x}{\log{\left(x \right)}}\) + C.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] The real domain is x>1. Let u=ln x, so dx=e^u du.
  • [1 mark] The integral becomes ∫e^u[ln u+u^(−2)]du. With g(u)=ln u−1/u, one has g+g′=ln u+u^(−2).
  • [1 mark] Use d[e^u g(u)]/du=e^u(g+g′); substitute u=ln x to get xln(ln x)−x/ln x.
  • [0.5 marks] Therefore the required integral is \(x \log{\left(\log{\left(x \right)} \right)} - \frac{x}{\log{\left(x \right)}}\) + C.

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