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Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 6 textbook solution.

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Answer : a

Given : \int_{-a}^{a} \frac{1}{1+a^{x}} d x, a>0

Hint : Use the formula of \int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x

Solution : I=\int_{-a}^{a} \frac{1}{1+a^{x}} d x \quad-----(1)

We know that \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x

\begin{aligned} &I=\int_{-a}^{a} \frac{1}{1+a^{-x}} d x \\ &I=\int_{-a}^{a} \frac{a^{x}}{a^{x}+1} d x-------(2) \end{aligned}

Add (1) and (2)

\begin{aligned} 2 I &=\int_{-a}^{a} \frac{1+a^{x}}{a^{x}+1} d x \\ 2 I &=\int_{-a}^{a} 1 . d x \end{aligned}

\begin{aligned} &2 I=(x)_{-a}^{a} \\ &2 I=2 a \\ &I=a \end{aligned}

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