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u=ln(x^2 + 1)-----)du=2x/(x^2+1) v=x--------------)dv= dx. note: der( u*v) = u dv+ v du. and then Integrate both sides and you get : uv= int(u dv)+ int (v du). Switch it around and you get int(u dv) =uv- int (v du) So the integral is ln(x^2 + 1)*x- int(2x^2/x^2+1) Next, integrate the...

i think this answer is: 1/4x(-x+2(x+2)lnx-4)+C

number 1-6 is my equation,...

This my data: 1.e^λx 2.e^iβx 3.sinαx 4.sin^2αx 5.cosαx x^2 How i find eigen value and eigen function from equation above?