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1/√(x)+1Dx=?

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=∫x^(-3/2)+1dx=-2x^(-1/2)+x+C=x-2/√x+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

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

let u=√x du = dx/(2√x) dx = 2u du ∫ dx/(√x -1) =∫ 2u/(u-1) du =∫ [2 + 2/(u-1)] du =2u +2ln|u-1| +C =2√x + 2|√x-1| +C ans : B

如图所示:

=∫(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

其中C为常数

您把第二个完全平方式展开,会发现两者区别仅在于常数。对于原函数而言,常数的区别算是区别吗?

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