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

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(B) We are given $y=	an ^{-1}\left[\frac{1}{1+x(1+x)}ight]+	an ^{-1}\left[\frac{1}{1+(x+2)(x+1)}ight]$.Using the formula $	an ^{-1} A - 	an ^{

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

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(B) We are given $y=	an ^{-1}\left[\frac{1}{1+x(1+x)}ight]+	an ^{-1}\left[\frac{1}{1+(x+2)(x+1)}ight]$.Using the formula $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1}\left(\frac{A-B}{1+AB}ight)$,we can rewrite the terms:$y = 	an ^{-1}\left(\frac{(x+1)-x}{1+(x+1)x}ight) + 	an ^{-1}\left(\frac{(x+2)-(x+1)}{1+(x+2)(x+1)}ight)$.This simplifies to:$y = (	an ^{-1}(x+1) - 	an ^{-1} x) + (	an ^{-1}(x+2) - 	an ^{-1}(x+1))$.Canceling the common terms,we get:$y = 	an ^{-1}(x+2) - 	an ^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(	an ^{-1}(x+2)) - \frac{d}{dx}(	an ^{-1} x)$.$\frac{dy}{dx} = \frac{1}{1+(x+2)^2} - \frac{1}{1+x^2}$.Thus,the correct option is $B$.

If $y=\tan ^{-1}\left[\frac{1}{1+x+x^2}\right]+\tan ^{-1}\left[\frac{1}{x^2+3 x+3}\right], x>0$,then $\frac{d y}{d x}=$

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