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  • Explain solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 25 Subquestion (i) maths textbook solution.

Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 25 Subquestion (i) maths textbook solution.     

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Answer:- 0

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

Given:-  \int_{-1}^{1} \log \left(\frac{2-x}{2+x}\right) d x

Solution :

\begin{aligned} &f(x)=\log \left(\frac{2-x}{2+x}\right) \\ &f(-x)=\log \left(\frac{2-x}{2+x}\right) \end{aligned}

              \begin{aligned} &=\log (2+x)-\log (2-x) \\ &=-[\log (2-x)-\log (2+x)] \\ &=-\log \left(\frac{2-x}{2+x}\right) \end{aligned}

\begin{aligned} &=-f(x) \\ f(-x) &=-f(x) \end{aligned}

f(x) is an odd function

So =\int_{-1}^{1} \log \left(\frac{2-x}{2+x}\right) \cdot d x=0

 

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