If $\displaystyle y = tan^{-1}\frac{1}{x^2+x+1} + tan^{-1}\frac{1}{x^2+3x+3} + tan^{-1} \frac{1}{x^2+5x+7} +.....$ to n terms. Then prove that $\displaystyle \frac{dy}{dx} = \frac{1}{(x+n)^2 + 1} - \frac{1}{x^2+1}$
If $\displaystyle y = tan^{-1}\frac{1}{x^2+x+1} + tan^{-1}\frac{1}{x^2+3x+3} + tan^{-1} \frac{1}{x^2+5x+7} +.....$ to n terms. Then prove that $\displaystyle \frac{dy}{dx} = \frac{1}{(x+n)^2 + 1} - \frac{1}{x^2+1}$
My hint would be by induction noting that the nth term in $\displaystyle y$ is
$\displaystyle \tan^{-1}\left( \frac{1}{x^2 +(2n-1)x + n^2-n+1}\right)$