If ∫tan⁡x1+tan⁡x+tan2⁡xdx=x−KAtan−1⁡Ktan⁡x+1A+C,(C is a constant of integration),  then the ordered pair (K,A) is equal to

Let I=∫tan⁡x1+tan⁡x+tan2⁡xdx⇒I=∫tan⁡x+1+tan2⁡xtan⁡x+1+tan2⁡xdx−∫1+tan2⁡x1+tan⁡x+tan2⁡xdx⇒I=x−∫sec2⁡x1+tan⁡x+tan2⁡xdx Put tan⁡x=t⇒sec2⁡x⋅dx=dt∴I=x−∫dtt2+t+1∴I=x−∫dtt2+t+14+1−14=x−∫dtt+122+322⇒I=x−23tan−1⁡t+1232+C⇒I=x−23tan−1⁡2tan⁡x+13+C∴A=3 and K=2