求原函数。
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原式=(1/3)∫arctanxd(x³)
=(1/3)[x³arctanx-∫x³d(arctanx)]
=(1/3)x³arctanx-(1/3)∫[x³/(1+x²)]dx
=(1/3)x³arctanx-(1/6)∫[x²/(1+x²)]d(x²)
=(1/3)x³arctanx-(1/6)∫[(x²+1-1)/(1+x²)]d(x²)
=(1/3)x³arctanx-(1/6)∫d(x²)+(1/6)∫[1/(1+x²)]d(x²+1)
=(1/3)x³arctanx-(1/6)x²+(1/6)ln(1+x²)+C
=(1/3)[x³arctanx-∫x³d(arctanx)]
=(1/3)x³arctanx-(1/3)∫[x³/(1+x²)]dx
=(1/3)x³arctanx-(1/6)∫[x²/(1+x²)]d(x²)
=(1/3)x³arctanx-(1/6)∫[(x²+1-1)/(1+x²)]d(x²)
=(1/3)x³arctanx-(1/6)∫d(x²)+(1/6)∫[1/(1+x²)]d(x²+1)
=(1/3)x³arctanx-(1/6)x²+(1/6)ln(1+x²)+C
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