这两个方法算出来答案不一样 10
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∫xarctan2xdx
=(1/2)∫arctan2xd(x²)
=(1/2)[x²*arctan2x-∫x²d(arctan2x)]
=(1/2)x²*arctan2x-(1/2)*∫x²[1/(1+4x²)]*2dx
=(1/2)x²*arctan2x-∫[x²/(1+4x²)]dx
=(1/2)x²*arctan2x-(1/4)∫[(4x²+1)-1]/(1+4x²)dx
=(1/2)x²*arctan2x-(1/4)x+(1/4)∫[1/(1+4x²)]dx
=(1/2)x²*arctan2x-(1/4)x+(1/8)∫[1/(1+4x²)]d(2x)
=(1/2)x²*arctan2x-(1/4)x+(1/8)arctan2x+C
=(1/2)∫arctan2xd(x²)
=(1/2)[x²*arctan2x-∫x²d(arctan2x)]
=(1/2)x²*arctan2x-(1/2)*∫x²[1/(1+4x²)]*2dx
=(1/2)x²*arctan2x-∫[x²/(1+4x²)]dx
=(1/2)x²*arctan2x-(1/4)∫[(4x²+1)-1]/(1+4x²)dx
=(1/2)x²*arctan2x-(1/4)x+(1/4)∫[1/(1+4x²)]dx
=(1/2)x²*arctan2x-(1/4)x+(1/8)∫[1/(1+4x²)]d(2x)
=(1/2)x²*arctan2x-(1/4)x+(1/8)arctan2x+C
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