If ∫2cosx−sinx+λcosx+sinx−2dx=Aln|cosx+sinx−2|+Bx+C then the value of λAB is
ddx(Aln|cosx+sinx−2|+Bx+C)=Acosx−sinxcosx+sinx−2+B=Acosx−Asinx+Bcosx+Bsinx−2Bcosx+sinx−2∴2=A+B,−1=−A+B,λ=−2B∴A=3/2,B=1/2,λ=−1∴λAB=−3/4