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∫(√x)^3%1/√x 1Dx

如图所示:

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

凑微分后代入上下限计算

=∫x^(-3/2)+1dx=-2x^(-1/2)+x+C=x-2/√x+C

余下就很容易了.

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

配方:1+x-x^2=5/4-(x-1/2)^2,套用不定积分公式(∫dx/√(a^2-x^2)) 结果是arcsin((2x-1)/√5)+C

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=∫(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

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