CBSE Class 12 Mathematics sample paper 2024-25: question 29(a)

Hints

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Step-by-step solution

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1. Split: $I=\int \frac{1}{\log x}dx-\int \frac{1}{(\log x{)}^2}dx$. Integrate the first by parts with $\frac{1}{\log x}$ as first function: $\int \frac{1}{\log x}dx=\frac{1}{\log x}\int dx-\int \left\{\frac{d}{dx}\left(\frac{1}{\log x}\right)\int dx\right\}dx$.

   1 mark

2. $=\frac{x}{\log x}-\int \left(-\frac{1}{(\log x{)}^2}\cdot \frac{1}{x}\right)xdx=\frac{x}{\log x}+\int \frac{1}{(\log x{)}^2}dx$.

   1 mark

3. So $I=\frac{x}{\log x}+\int \frac{dx}{(\log x{)}^2}-\int \frac{dx}{(\log x{)}^2}=\frac{x}{\log x}+C$.

   1 mark

Answer: $\frac{x}{\log x}+C$

Revise the chapter

• Chapter: Integrals (notes, formulas, where marks are lost and every past question)