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  • Provide solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 34 textbook solution.

Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 34 textbook solution.

Answers (1)

Answer:-  0

Hints:-  You must know about the integration rules of logarithm functions.

Given:-   I=\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x

Solution :

\begin{aligned} & I=\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x \\ &I=\int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x \end{aligned}                                                     ....(1)

\begin{aligned} &I=\int_{0}^{1} \log \frac{(1-(1-x))}{1-x} d x \\ &I=\int_{0}^{1} \log \frac{x}{1-x} d x \end{aligned}                                            .....(2)

Adding both

     \begin{aligned} &2 I=\int_{0}^{1} \log \frac{1-x}{x} d x+\int_{0}^{1} \log \frac{x}{1-x} d x \\ &2 I=\int_{0}^{1} \log \left(\frac{1-x}{x} \cdot \frac{x}{1-x}\right) d x \\ &2 I=\int_{0}^{1} \log 1 \cdot d x \end{aligned}

     \begin{aligned} &2 I=0 \\ &I=0 \end{aligned}

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