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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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