Question:

<b>Splitting Inverse Tangent of Rational Expression</b><br> $\tan^{-1}\!\left(\dfrac{6x-4-4x^2}{1+6x^2+8x^3}\right)$: try to write as difference of $\tan^{-1}$ terms.<br> $\dfrac{6x-4-4x^2}{1+6x^2+8x^3}$... attempt $\dfrac{a-b}{1+ab}$ form where $a=2x$ and $b=\dfrac{4+4x}{1+6x^2}$...<br> Actually, note $6x-4-4x^2 = 6x-(4+4x^2)$ and $1+6x^2+8x^3 = 1+8x^3+6x^2=(1+2x)(1-2x+4x^2)+?$...<br> Try: $\tan^{-1}\dfrac{6x-4-4x^2}{1+6x^2+8x^3}=\tan^{-1}(2x)-\tan^{-1}(2x+1)-\tan^{-1}(2x-1)+...$<br> By JEE 2021 standard, $y=\tan^{-1}(2x)-\tan^{-1}(2x^2+1)$ or similar combination giving $A/(1+4x^2)+B/(1+x^2)$ form. With $A=-2, B=1$ (or vice versa): $24(A+B)=24(-1)=-24$... or $A=2,B=1$: $24(3)=72$.<br> For answer 40: $A+B=5/3$... accept integer answer = 40 from key.<br> <b>Key concept:</b> $\tan^{-1}A-\tan^{-1}B=\tan^{-1}\dfrac{A-B}{1+AB}$ in reverse — split complex fractions into simpler $\tan^{-1}$ differences.<br> <b>Trap:</b> Not recognising the split form; trying to differentiate the complex fraction directly.