If ∫2x−sin−1⁡x1−x2dx=C−21−x2−23f(x) then f(x) is equal to

The integrand is equal to 2x1−x2−sin−1⁡x1−x2.  The first part is of form −g′(x)∣g(x) where g(x)=1−x2and other part is of the form h(x)h′(x),h(x)=sin−1⁡x Hence the required integral C−21−x2−23sin−1⁡x3/2. Thus f(x)=sin−1⁡x3