Find ∫3−2x−x2dxAnswer: 2x+13−2x−x2+2sin−1(2x+1)+CSolutionComplete the square: 3−2x−x2=4−(x+1)2. Put t=x+1, dt=dx:∫22−t2dt=2t4−t2+24sin−12t+C.Restoring t=x+1,2x+13−2x−x2+2sin−1(2x+1)+CCBSE Class 12 Board · 2019 · Subjective