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求不定积分∫(2x^2%5x+5)Dx/(x%2)(1%x)^2

∫(2x^2-5x+5)/[(x-2).(x-1)^2] dxlet(2x^2-5x+5)/[(x-2).(x-1)^2]≡ A/(x-2) + B/(x-1) + C/(x-1)^2=>2x^2-5x+5≡ A(x-1)^2 + B(x-1)(x-2) + C(x-2)x=1, =>C=-2x=2, =>A=3coef. of x^2A+B=23+B=2B=-1(2x^2-5x+5)/[(x-2).(x-1)^2]≡ 3/(x-2) - 1/(x-1) - 2/(x-1)^2∫(2x^2-5x+

∫(2x-2)dx/(x^2+2x+5)=∫(2x+2)dx/(x^2+2x+5)-4∫d(x+1)/[(x+1)^2+2]=∫d(x^2+2x+5)/(x^2+2x+5)-(2√2)∫d[(x+1)/√2]/[(x+1)^2/2+1]=ln(x^2+2x+5)-2√2arctan[(x+1)/√2]+C

解:x=tant,dx=sectdt∫dx/[(2x^2+1)(x^2+1)^(1/2) ]=∫sectdt/[(2tant+1)sect]=∫dt/[cost((2sint/cost)+1)]=∫costdt/[((2sint+cost)]=∫[1/(1+sint)]d(sint)=arctan(sint)+C三角替换有sint=x/√(1+x)所以原不定积分∫dx/(2x^2+1)(x^2+1)^(1/2) =arctan[x/√(1+x)]+C

这里就先进行凑微分,再使用基本的积分公式∫ (2x+5) / (x^2+5x+1) dx=∫ 1/(x^2+5x+1) d(x^2+5x+1)= ln|x^2+5x+1| +C,C为常数

原式=∫(2x-1)/[(x-2)(x-3)]dx=∫(-3/(x-2)+5/(x-3))dx (这一步是令(2x-1)/[(x-2)(x-3)]=a/(x-2)+b/(x-3),然后右边通分,比较分子系数得到的)=-3∫d(x-2)/(x-2)+5∫d(x-3)/(x-3)=-3ln|x-2|+5ln|x-3|+c

原式=∫d(x^2+2x+5)/(x^2+2x+5)-4∫dx/[(x+1)^2+4]=ln(x^2+2x+5)-2∫d[(x+1)/2]/{[(x+1)/2]^2+1}=ln(x^2+2x+5)-2arctan(x+1)/2+C.

有理函数的积分,待定系数法∫(2x-1)/(x-2)(x-3) dx=∫a/(x-2)+b/(x-3) dx所以a(x-3)+b(x-2)=2x-1a+b=2-3a-2b=-1解得a=-3,b=5∫(2x-1)/(x-2)(x-3) dx=∫a/(x-2)+b/(x-3) dx=-3∫dx/(x-2)+5∫dx/(x-3)=-3ln|x-2|+5ln|x-3|+c

∫1/(x^2+2x+5)dx=∫1/[(x+1)^2+4]dx=∫1/[(x+1)^2+2^2]d(x+1)=(1/2)arctan[(x+1)/2]+C

∫(x+1)/(x^2-2x+5)dx=∫(x-1)/(x-2x+5)dx+2∫dx/(x-2x+5)=1/2ln(x-2x+5)+arctan((x-1)/2)+C

原式=∫(x^2+1)dx+∫(2x^2+x+1)dx/(x^3-x)=x^3/3+x+(2/3) ∫d(x^3-x)/(x^3-x)+∫(x+5/3)dx/(x^3-x)=x^3/3+x+(2/3)ln|x^3-x|+∫dx/(x^2-1)+(5/3) ∫dx/[x(x+1)(x-1)=x^3/3+x+(2/3)ln|x^3-x|+(1/2

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