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∫1/1+³√x+1Dx

换元令x=t^6,t=x^(1/6) =∫(t³-1)/(t²+1)dt^6 =6∫(t^8-t^5)/(t²+1)dt =6∫t^6-t^4-t³+t²+t-1-t/(t²+1)+1/(t²+1)dt =6t^7/7-6t^5/5-3t^4/2+2t³+3t²-6t-3ln(t²+1)+6arctant+C

=∫(x+1/2)/(x²+x+1)dx+1/2∫1/((x+1/2)²+3/4)dx =1/2∫1/(x²+x+1)d(x²+x+1)+1/2∫1/(u²+3/4)du =(1/2)ln(x²+x+1)+(1/2)/(3/4)*√3/2*arctan(2u/√3)+C =(1/2)ln(x²+x+1)+(1/√3)arctan((2x+1)/√3)+C

原式=∫(x-1)/(x²+1)dx =∫xdx/(x²+1)-∫dx/(x²+1) =∫0.5d(x²)/(x²+1)-arctanx =0.5ln(x²+1)-arctanx+C

=∫(1+1/x²)/(x²+1/x²)dx =∫1/(x²+1/x²)d(x-1/x) 换元t=x-1/x =∫1/(t²+2)dt =(1/√2)arctan(t/√2)+C 然后代入t=x-1/x

∫[(1-x)/(x²+1)]dx =∫[1/(x²+1)]dx- ½∫[1/(x²+1)]d(x²+1) =arctanx -½ln(x²+1) +C

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