Find ∫4x−x2dxAnswer: sin−1(2x−2)+CSolutionComplete the square: 4x−x2=4−(x−2)2=22−(x−2)2. Then∫22−(x−2)2dx=sin−1(2x−2)+C.sin−1(2x−2)+CCBSE Class 12 Board · 2022 · Subjective