求不定积分3-x^2/x^3-xdx
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求不定积分3-x^2/x^3-xdx
求不定积分3-x^2/x^3-xdx
求不定积分3-x^2/x^3-xdx
思路:利用待定系数法
x^3-x=x(x-1)(x+1)
设3-x^2/x^3= a/x +b/(x-1) +c/(x+1)=[(a+b+c)x²+(b-c)x-a]/[x(x-1)(x+1)]
所以有
a+b+c=-1
b-c=0
-a=3
解得:a=-3 n=c=1
所以
3-x^2/x^3= -3/x +1/(x-1) +1/(x+1)
三项分别积分.
思路:利用待定系数法
x^3-x=x(x-1)(x+1)
设3-x^2/x^3= a/x +b/(x-1) +c/(x+1)=[(a+b+c)x²+(b-c)x-a]/[x(x-1)(x+1)]
所以有
a+b+c=-1
b-c=0
-a=3
解得:a=-3 n=c=1
所以
3-x^2/x^3= -3/x +1/(x-1) +1/(x+1)
三项分别积分。
有两种,你选一种吧!∫[(3-x^2)/(x^3-x)]dx
=(1/2)∫[(2-x^2+1)/(x^4-x^2)]d(x^2)
=(1/2)∫[2/(x^4-x^2)]d(x^2)-(1/2)∫[(x^2-1)/(x^4-x^2)]d(x^2)
=∫[1/(x^4-x^2)]d(x^2)-(1/2)∫(1/x^2)d(x^2)
=∫[(x^2-x^2+1)/(x...
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有两种,你选一种吧!∫[(3-x^2)/(x^3-x)]dx
=(1/2)∫[(2-x^2+1)/(x^4-x^2)]d(x^2)
=(1/2)∫[2/(x^4-x^2)]d(x^2)-(1/2)∫[(x^2-1)/(x^4-x^2)]d(x^2)
=∫[1/(x^4-x^2)]d(x^2)-(1/2)∫(1/x^2)d(x^2)
=∫[(x^2-x^2+1)/(x^4-x^2)]d(x^2)-(1/2)ln(x^2)
=∫[x^2/(x^4-x^2)]d(x^2)-∫[(x^2-1)/(x^4-x^2)]d(x^2)-ln|x|
=∫[1/(x^2-1)]d(x^2)-∫(1/x^2)d(x^2)-ln|x|
=ln|x^2-1|-ln(x^2)-ln|x|+C
=ln|x^2-1|-3ln|x|+C。 思路:利用待定系数法
x^3-x=x(x-1)(x+1)
设3-x^2/x^3= a/x +b/(x-1) +c/(x+1)=[(a+b+c)x²+(b-c)x-a]/[x(x-1)(x+1)]
所以有
a+b+c=-1
b-c=0
-a=3
解得:a=-3 n=c=1
所以
3-x^2/x^3= -3/x +1/(x-1) +1/(x+1)
三项分别积分。
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