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1,∫3/x^3+1Dx 2,∫Dx/x(x^6+4)

sin^2(x)cos^4(x) =1/4*sin²2xcos²x =1/4*(1-cos4x)/2*(1+cos2x)/2 =1/16*(1+cos2x-cos4x-cos2xcos4x) =1/16*(1+cos2x-cos4x-cos2xcos4x) =1/16*[1+cos2x-cos4x-1/2*(cos3x+cos6x)] =1/16+1/16*cos2x-1/16*cos4x-1/32*cos3x-1/32*cos6...

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