If y = tan^{-1}left(frac{1}{x^2 + x + 1}right | vedclass

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(A) The general term of the series is $T_r = 	an^{-1}\left(\frac{1}{x^2 + (2r-1)x + (r^2-r+1)}ight)$ for $r = 1, 2, \dots, n$. This can be written

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$\frac{1}{1 + (x + n)^2} - \frac{1}{1 + x^2}$

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(A) The general term of the series is $T_r = 	an^{-1}\left(\frac{1}{x^2 + (2r-1)x + (r^2-r+1)}ight)$ for $r = 1, 2, \dots, n$. This can be written as $T_r = 	an^{-1}\left(\frac{(x+r) - (x+r-1)}{1 + (x+r)(x+r-1)}ight)$. Using the formula $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,we get $T_r = 	an^{-1}(x+r) - 	an^{-1}(x+r-1)$. Summing up to $n$ terms: $y = [	an^{-1}(x+1) - 	an^{-1} x] + [	an^{-1}(x+2) - 	an^{-1}(x+1)] + \dots + [	an^{-1}(x+n) - 	an^{-1}(x+n-1)]$. This is a telescoping series,so $y = 	an^{-1}(x+n) - 	an^{-1} x$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{1 + (x+n)^2} - \frac{1}{1 + x^2}$.

If $y = \tan^{-1}\left(\frac{1}{x^2 + x + 1}\right) + \tan^{-1}\left(\frac{1}{x^2 + 3x + 3}\right) + \tan^{-1}\left(\frac{1}{x^2 + 5x + 7}\right) + \dots$ up to $n$ terms,then $\frac{dy}{dx}$ is equal to

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