If ∫tanx1+tanx+tan2xdx=x−KAtan−1Ktanx+1A+C,(C is a constant of integration), then the ordered pair (K,A) is equal to
2,3
2,1
−2,1
−2,3
Let I=∫tanx1+tanx+tan2xdx⇒I=∫tanx+1+tan2xtanx+1+tan2xdx−∫1+tan2x1+tanx+tan2xdx⇒I=x−∫sec2x1+tanx+tan2xdx Put tanx=t⇒sec2x⋅dx=dt∴I=x−∫dtt2+t+1
∴I=x−∫dtt2+t+14+1−14=x−∫dtt+122+322⇒I=x−23tan−1t+1232+C⇒I=x−23tan−12tanx+13+C∴A=3 and K=2