不定积分~1∫dx/(x^2+1
解：1。谈不上简单，只能说是经验
1/[(x²+1)(x²+x+1)]
=(1/x)[1/(x²+1)-1/(x²+x+1)]
=1/[x(x²+1)]-1/[x(x²+x+1)]
=(1/x)-x/(x²+1)-[(1/x)-(x+1)/(x²+x+1)]
=(x+1)/(x²+x+1)-x/(x²+1)
故原积分=∫(x+1)dx/(x²+x+1)-∫xdx/(x²+1)
=∫(x+1/2)dx/(x²+x+1)+(1/2)∫dx/(x²+x+1)...全部
解：1。谈不上简单，只能说是经验
1/[(x²+1)(x²+x+1)]
=(1/x)[1/(x²+1)-1/(x²+x+1)]
=1/[x(x²+1)]-1/[x(x²+x+1)]
=(1/x)-x/(x²+1)-[(1/x)-(x+1)/(x²+x+1)]
=(x+1)/(x²+x+1)-x/(x²+1)
故原积分=∫(x+1)dx/(x²+x+1)-∫xdx/(x²+1)
=∫(x+1/2)dx/(x²+x+1)+(1/2)∫dx/(x²+x+1)
-(1/2)∫dx²/(x²+1)
=(1/2)∫d(x²+x+1)/(x²+x+1)+(1/2)∫dx/[(x+1/2)²+(3/4)]
-(1/2)∫d(x²+1)/(x²+1)
=(1/2)ln(x²+x+1)+(1/2)arctan[(x+1/2)/(√3/2)]
-(1/2)ln(x²+1)+C
=(1/2)ln(x²+x+1)+(1/2)arctan(2x+1/√3)-(1/2)ln(x²+1)+C
2。
要熟悉∫(1+x²)dx/(x^4+1)和∫(1-x²)dx/(x^4+1)的积分方法才能想到用下面的求法
∫dx/(x^4+1)
=(1/2)∫[(1+x²)+(1-x²)]dx/(x^4+1)
=(1/2)∫(1+x²)dx/(x^4+1)+(1/2)∫(1-x²)dx/(x^4+1)
=(1/2)∫[1+(1/x²)]dx/[x²+(1/x²)](分子分母同除以x²)
-(1/2)∫[1-(1/x²)]dx/[x²+(1/x²)]
=(1/2)∫d[x-(1/x)]/{[x-(1/x)]²+2}
-(1/2)∫d[x+(1/x)]/{[x+(1/x)]²-2}
=(1/2√2)arctan{[x-(1/x)]/√2}
-(1/4√2)ln|[x+(1/x)-√2]/[x+(1/x)+√2]|+C
=(√2/4)arctan[(x²-1)/x√2}
-(√2/8)ln|(x²-x√2+1)/(x²+x√2+1)|+C。
收起
