amp;I(x)=∫(x−11)1311(x+15)1315dxamp; Put x+15x−11=t⇒(x+5)226dx=dtamp;I(x)=261∫t11/13dt=261⋅2/13t2/13
amp;I(x)=41(x+15x−11)2/13+Camp;I(37)−I(24)=41(5226)2/13−41(3913)2/13amp;=41(22/131−32/131)amp;=41(41/131−91/131)amp;∴b=4,c=9amp;3( b+c)=39