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Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 36

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Answer:  0

Hint: To solve the equation, the value of negative is always zero

Given:  \int_{-a}^{a} \frac{x e^{x^{2}}}{1+x^{2}} d x


\begin{aligned} &\int_{-a}^{a} f(x) d x=\int_{-a}^{a} f(a-a-x) d x \\ & \end{aligned}

=\int_{-a}^{a} f(-x) d x

\begin{aligned} &=-\int_{-a}^{a} f(x) d x \\ & \end{aligned}

I=\int_{-a}^{a} \frac{(-x) e^{x^{2}}}{1+x^{2}} d x \\

2 I=0 \\


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