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

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(D) Given that $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$.We know that $\int \frac{dx}{1+x^2} = 	an^{-1}(x) + C$.Applying the lim

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(D) Given that $\int_0^b \frac{dx}{1+x^2} = \int_b^{\infty} \frac{dx}{1+x^2}$.We know that $\int \frac{dx}{1+x^2} = 	an^{-1}(x) + C$.Applying the limits:$[	an^{-1}(x)]_0^b = [	an^{-1}(x)]_b^{\infty}$$	an^{-1}(b) - 	an^{-1}(0) = \lim_{x 	o \infty} 	an^{-1}(x) - 	an^{-1}(b)$Since $	an^{-1}(0) = 0$ and $\lim_{x 	o \infty} 	an^{-1}(x) = \frac{\pi}{2}$, we have:$	an^{-1}(b) - 0 = \frac{\pi}{2} - 	an^{-1}(b)$$2 	an^{-1}(b) = \frac{\pi}{2}$$	an^{-1}(b) = \frac{\pi}{4}$$b = 	an\left(\frac{\pi}{4}ight)$$b = 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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