The value of int_{-1}^{3} left( tan^{-1} left | vedclass

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Question

(A) We know that $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for any $u > 0$. Also,$	an^{-1} \left( \frac{1}{u} ight) = \cot^{-1}(u)$ for $u > 0$

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$2\pi$

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(A) We know that $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for any $u > 0$. Also,$	an^{-1} \left( \frac{1}{u} ight) = \cot^{-1}(u)$ for $u > 0$. Let $I = \int_{-1}^{3} \left( 	an^{-1} \left( \frac{x}{x^2+1} ight) + 	an^{-1} \left( \frac{x^2+1}{x} ight) ight) dx$. Note that the integrand is defined for $x 
eq 0$. Using the property $	an^{-1} \left( \frac{1}{u} ight) = \cot^{-1}(u)$,the expression becomes: $I = \int_{-1}^{3} \left( 	an^{-1} \left( \frac{x}{x^2+1} ight) + \cot^{-1} \left( \frac{x}{x^2+1} ight) ight) dx$. Since $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$,the integrand simplifies to $\frac{\pi}{2}$. Thus,$I = \int_{-1}^{3} \frac{\pi}{2} dx = \frac{\pi}{2} [x]_{-1}^{3} = \frac{\pi}{2} (3 - (-1)) = \frac{\pi}{2} (4) = 2\pi$.

The value of $\int_{-1}^{3} \left( \tan^{-1} \left( \frac{x}{x^2+1} \right) + \tan^{-1} \left( \frac{x^2+1}{x} \right) \right) dx$ is

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