The integral int_{-1}^{3} left( tan^{-1} frac | vedclass

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(B) Let $I = \int_{-1}^{3} \left( 	an^{-1} \frac{x}{x^2+1} + 	an^{-1} \frac{x^2+1}{x} ight) dx$. Using the identity $	an^{-1}(u) + \cot^{-1}(u) =

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$2\pi$

Vedclass · Answer

(B) Let $I = \int_{-1}^{3} \left( 	an^{-1} \frac{x}{x^2+1} + 	an^{-1} \frac{x^2+1}{x} ight) dx$. Using the identity $	an^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for $u > 0$,we note that $	an^{-1} \frac{x^2+1}{x} = \cot^{-1} \frac{x}{x^2+1}$. Thus,the integrand becomes $	an^{-1} \left( \frac{x}{x^2+1} ight) + \cot^{-1} \left( \frac{x}{x^2+1} ight) = \frac{\pi}{2}$. Therefore,$I = \int_{-1}^{3} \frac{\pi}{2} dx$. $I = \frac{\pi}{2} [x]_{-1}^{3} = \frac{\pi}{2} (3 - (-1)) = \frac{\pi}{2} (4) = 2\pi$.

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

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