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x的4次方/根号下(1%x的2次方)的不定积分是多少

令 x = sint,I = ∫x^4dx/√(1-x^2) = ∫(sint)^4 costdx/cost = ∫(sint)^4 dx= (1/4)∫(1-cos2t)^2 dt = (1/4)∫[1-2cos2t+(cos2t)^2] dt= (1/4)∫[1-2cos2t+(1/2)(1+cos4t)] dt= (1/4)∫[3/2-2cos2t+(1/2)cos4t] dt= (1/4)[3t/2 - sin2t + (1/8)sin4t] + C= (1/4)[3t/2 - 2sintcost +

结果是 (1/2)[arcsinx + x√(1 - x)] + C x = sinθ,dx = cosθ dθ ∫ √(1 - x) dx = ∫ √(1 - sinθ)(cosθ dθ) = ∫ cosθ dθ= ∫ (1 + cos2θ)/2 dθ = θ/2 + (sin2θ)/4 + C= (arcsinx)/2 + (sinθcosθ)/2 + C= (arcsinx)/2 + (x√(1 - x))/2 + C= (1/2)[arcsinx + x√(1

假设x^4+1=(x^2+Ax+B)(x^2+Cx+D)用待定系数法可得 A=√2,C=-√2,B=D=1 然后再根据拆解分式的方法原式必可表示成1/(x^4+1)=(ax+b)/(x^2+√2x+1)+(cx+d)/(x^2-√2x+1)+e 用待定系数法得到 a=√2/4,c=-√2/4,b=d=1/2,e=0 下面其实就好求了

∫x^2dx/√(1-x^4)=∫1dx/√((1/x^4)-1)=(1/2)(∫(√(1/x^2)+1)dx/(√(1/x^2)-1)-∫(√(1/x^2)-1)dx/(√(1/x^2)+1))=(1/2)(∫2dx/(√(1/x^2)-1)-∫1dx-∫2dx/(√(1/x^2)+1)+∫1dx)=∫1dx/(√(1/x^2

解题过程如下:1/√(1+x^4)=(1+x^4)^(-1/2)=1-(1/2)x^4+(-1/2)(-1/2-1)/2!x^8+…+(-1/2)(-1/2-1)…(-1/2-n+1)/n!x^(4n)+…=1+∑(n:1→∞)(-1)^n(2n-1)!/(2n)!x^(4n),x∈(-1,1) ∫1/√(1+x^4)dx=x+∑(n:1→∞)(-1)^n(2n-1)!/[(4n+1)(2n)!]x^(4n+1)+

令x=sinu,则:u=arcsinx,dx=cosudu.∴∫{x^4/√[(1-x^2)^3]}dx=∫[(sinu)^4/(cosu)^3]cosudu=∫{[1-(cosu)^2]^2/(cosu)^2}du=∫[1/(cosu)^2]du-2∫du+∫(cosu)^2du=tanu-2u+(1/2)∫(1+cos2u)du=tanu-2u+(1/2)∫du+(1/4)∫cos2ud(2u)=sinu/cosu-(1/2)arcsinx+(1/4)sin2u+C=x/√(1-x^2)-(1/2)arcsinx+(1/2)sinucosu+C=x/√(1-x^2)-(1/2)arcsinx+(1/2)x√(1-x^2)+C.

∫x^4/(1+x^2)dx=∫(x^4-1+1)/(1+x^2)dx=∫[x^2-1+1/(1+x^2)]dx=x^3/3-x+arctanx+C

原式=∫1\(2^2-x^2)dx=-∫1\(x^2-2^2)dx =-1\4(ln|(x-2)\(x+2)|+C

∫xdx/√(4 - x)= (1/2)∫dx/√(4 - x)= -(1/2)∫d(4 - x)/√(4 - x)= -(1/2)[1/(-1/2 + 1)]√(4 - x) + C= -√(4 - x) +C

令x = 2sinθ,dx = 2cosθdθ∫ x/√(4 - x) dx= ∫ 4sinθ/√(4 - 4sinθ) * (2cosθdθ)= ∫ 4sinθ/|2cosθ| * (2cosθdθ)= 4∫ sinθ dθ= 2∫ (1 - cos2θ) dθ= 2θ - 2(1/2)sin2θ + C= 2θ - 2sinθcosθ + C= 2arcsin(x/2) - 2(x/2)√(4 - x)/2 + C= 2arcsin(x/2) - (x/2)√(4 - x) + C

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