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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.     

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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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