<s>[INST] $y=\tan^{-1}\frac{1}{x^2+x+1}+\tan^{-1}\frac{1}{x^2+3x+3}+\tan^{-1}\frac{1}{x^2+5x+7}+\tan^{-1}\frac{1}{x^2+7x+13}......$to n terms. Prove that if $y=\tan^{-1}\frac{1}{x^2+x+1}+\tan^{-1}\frac{1}{x^2+3x+3}+\tan^{-1}\frac{1}{x^2+5x+7}+\tan^{-1}\frac{1}{x^2+7x+13}......$to n terms.Then $\frac{dy}{dx}=\frac{1}{1+(x+n)^2}-\frac{1}{1+x^2}$
I could simplify only first term of $y$,not rest.
$\tan^{-1}\frac{1}{x^2+x+1}=\tan^{-1}\frac{(x+1)-x}{1+x(x+1)}=\tan^{-1}x+1-\tan^{-1}x$
and i could not judge $n$th term in $y$.Please help me. 
 [/INST] We can write $$\displaystyle \tan^{-1}\left(\frac{1}{x^2+x+1}\right) = \tan^{-1}\left(\frac{1}{1+(x+1)\cdot x}\right) = \tan^{-1}\left(x+1\right)-\tan^{-1}(x)$$.
Similarly $$\displaystyle \tan^{-1}\left(\frac{1}{x^2+3x+3}\right) = \tan^{-1}\left(\frac{1}{1+(x+2)\cdot (x+1)}\right) = \tan^{-1}\left(x+2\right)-\tan^{-1}(x+1)$$.
Similarly $$\displaystyle \tan^{-1}\left(\frac{1}{x^2+5x+7}\right) = \tan^{-1}\left(\frac{1}{1+(x+3)\cdot (x+2)}\right) = \tan^{-1}\left(x+3\right)-\tan^{-1}(x+2)$$.
Similarly for $\bf{n^{th}}$ term
$$\displaystyle \tan^{-1}\left(\frac{1}{x^2+x(2n-1)+n^2-n+1}\right) = \tan^{-1}\left(\frac{1}{1+(x+n)\cdot (x+n-1)}\right) = \tan^{-1}\left(x+n\right)-\tan^{-1}(x+n-1)$$.
Now Adding all these, we get $$\displaystyle y = \tan^{-1}(x+n)-\tan^{-1}(x)$$
And Differentitae both side w . r . to $x\;,$ We get $$\displaystyle \frac{dy}{dx} = \frac{1}{1+(x+n)^2}-\frac{1}{1+x^2}.$$
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