The value of int_{-2}^{2} left[ p ln left( fr | vedclass

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(C) Let $f(x) = p \ln \left( \frac{1+x}{1-x} \right) + q \ln \left( \frac{1-x}{1+x} \right)^{-2} + r$. Note that $\ln \left( \frac{1-x}{1+x} \right)^{

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The value of $r$

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(C) Let $f(x) = p \ln \left( \frac{1+x}{1-x} \right) + q \ln \left( \frac{1-x}{1+x} \right)^{-2} + r$. Note that $\ln \left( \frac{1-x}{1+x} \right)^{-2} = -2 \ln \left( \frac{1-x}{1+x} \right) = 2 \ln \left( \frac{1+x}{1-x} \right)$. Thus,$f(x) = (p + 2q) \ln \left( \frac{1+x}{1-x} \right) + r$. Since $\ln \left( \frac{1+x}{1-x} \right)$ is an odd function,the integral of this term over the symmetric interval $[-2, 2]$ is $0$. Therefore,$\int_{-2}^{2} f(x) dx = \int_{-2}^{2} r dx = r [x]_{-2}^{2} = r(2 - (-2)) = 4r$. Thus,the value of the integral depends only on $r$.

The value of $\int_{-2}^{2} \left[ p \ln \left( \frac{1+x}{1-x} \right) + q \ln \left( \frac{1-x}{1+x} \right)^{-2} + r \right] dx$ depends on:

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