If ∫2cos⁡x−sin⁡x+λcos⁡x+sin⁡x−2dx=Aln⁡|cos⁡x+sin⁡x−2|+Bx+C then the value of λAB is

ddx(Aln⁡|cos⁡x+sin⁡x−2|+Bx+C)=Acos⁡x−sin⁡xcos⁡x+sin⁡x−2+B=Acos⁡x−Asin⁡x+Bcos⁡x+Bsin⁡x−2Bcos⁡x+sin⁡x−2∴2=A+B,−1=−A+B,λ=−2B∴A=3/2,B=1/2,λ=−1∴λAB=−3/4