The correct option is
A 12ln∣∣∣tanxtanx+2∣∣∣+CI=∫dxsin2x+sin2x
Multiply numerator and denominator by sec2x
I=∫sec2xdxtan2x+2tanx
[Substitute tanx=t]⟹sec2xdx=dt
I=∫dtt2+2t
=12∫(1t−1t+2)dt
=12[ln(t)−ln(t+2)]+C
=12[ln(tanx)−ln(tanx+2)]+C
=12ln∣∣∣tanxtanx+2∣∣∣+C