If int_0^b frac{dx}{1+x^2} = int_b^{infty} fr | vedclass

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(D) We have,$\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$.Integrating both sides,we get $[	an^{-1} x]_0^b = [	an^{-1} x]_b^{\infty}

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$1$

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(D) We have,$\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$.Integrating both sides,we get $[	an^{-1} x]_0^b = [	an^{-1} x]_b^{\infty}$.Substituting the limits,$	an^{-1}(b) - 	an^{-1}(0) = 	an^{-1}(\infty) - 	an^{-1}(b)$.Since $	an^{-1}(0) = 0$ and $	an^{-1}(\infty) = \frac{\pi}{2}$,we have $	an^{-1}(b) = \frac{\pi}{2} - 	an^{-1}(b)$.Adding $	an^{-1}(b)$ to both sides,we get $2 	an^{-1}(b) = \frac{\pi}{2}$.Therefore,$	an^{-1}(b) = \frac{\pi}{4}$.Taking tangent on both sides,$b = 	an(\frac{\pi}{4}) = 1$.

If $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$,then $b$ is equal to

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