1樓:明哥歸來
∫(x+3)/(x2+5x+6)dx
=∫(x+3)/(x+2)(x=3)dx
=∫1/(x+2)dx
=ln|x+2|+c
怎樣求(x^3-3)/(5x-x^3)的不定積分
2樓:匿名使用者
x^3-3
=-(5x-x^3)+5x -3
∫(x^3-3)/(5x-x^3) dx
=-∫dx +5∫dx/(5-x^2) - 3∫dx/(5x-x^3)
=-x -5∫dx/(x^2-5) + 3∫dx/[x(x^2-5)]
=-x - (√5/2)∫ [1/(x-√5) -1/(x+√5) ]dx + 3∫dx/[x(x^2-5)]
=-x - (√5/2)ln|62616964757a686964616fe59b9ee7ad9431333339653766(x-√5)/(x+√5)| + 3∫dx/[x(x^2-5)]
=-x - (√5/2)ln|(x-√5)/(x+√5)| + 3[-(1/5)ln|x| + (1/10)ln|x^2-5|] +c
let1/[x(x^2-5)]≡ a/x + (bx+c)/(x^2-5)
=>1≡ a(x^2-5)+x(bx+c)
coef. of constant => a=-1/5
coef. of x => c=0
coef. of x^2
a+b=0
b=1/5
1/[x(x^2-5)]≡ -(1/5)(1/x) + (1/5)[x/(x^2-5)]
∫dx/[x(x^2-5)]
=-(1/5)∫(1/x)dx + (1/5)∫[x/(x^2-5)] dx
=-(1/5)ln|x| + (1/10)∫[2x/(x^2-5)] dx
=-(1/5)ln|x| + (1/10)ln|x^2-5| +c'
3樓:匿名使用者
^^設(x^3-3)/(5x-x^3)
=-1+(-5x+3)/[x(x-√5)(x+√5)]=-1+a/x+b/(x-√5)+c/(x+√5),則-5x+3=a(x^2-5)+x[b(x+√5)+c(x-√5)]=(a+b+c)x^2+√5(b-c)x-5a,比較係數得內a+b+c=0,b-c=-√5,a=-3/5,解得b=(3/5-√5)/2,c=(3/5+√5)/2,∴∫容(x^3-3)dx/(5x-x^3)=∫dx
=-x-(3/5)lnx+[(3/5-√5)/2]ln(x-√5)+[(3/5+√5)/2]ln(x+√5)+c.
x^2-5x+9/x^2-5x+6dx求不定積分
4樓:
∫(x2-5x)cosxdx
=∫(x2-5x)dsinx
=(x2-5x)sinx-∫sinxd(x2-5x)=(x2-5x)sinx-∫(2x-5)sinxdx=(x2-5x)sinx+∫(2x-5)dcosx=(x2-5x)sinx+(2x-5)cosx-∫cosxd(2x-5)
=(x2-5x)sinx+(2x-5)cosx-2∫cosxdx=(x2-5x)sinx+(2x-5)cosx-2sinx+c
求不定積分∫(x+3)/(x2+5x+6)dx
5樓:匿名使用者
∫(x+3)/(x2+5x+6)dx
=∫(x+3)/(x+2)(x=3)dx
=∫1/(x+2)dx
=ln|x+2|+c
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