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求∫x^2 Dx/(A^2+x^2)^(3/2)的不定积分

令x = a tanθ,dx = a secθ dθ(a + x)^(3/2) = (a + a tanθ)^(3/2) = (a secθ)^(3/2) = asecθ ∫ x/(a + x)^(3/2) dx= ∫ (atanθ)(asecθ)/(asecθ) dθ= ∫ tanθ/secθ dθ = ∫ (1 - cosθ)/cosθ dθ = ∫ (secθ - cosθ) dθ

∫dx/(x+a)^3/2,x=atant,dx=asectdt=a∫[sect/(atant+a)^3/2]dt=a∫[sect/asect]dt=(1/a)∫cosdt=(1/a)*sint+C=(1/a)*[x/√(x+a)]+C,C为任意常数

令 x = a tant, dx = a (sect)^2 dt, (a^2+x^2)^(3/2) = a^3 (sect)^3原式= ∫ a^(-2) cost dt = a^(-2) sint + C = a^(-2) x /[(a^2+x^2)^(1/2)]+ C

令x = a tanθ,dx = a secθ dθ(a + x)^(3/2) = (a + a tanθ)^(3/2) = (a secθ)^(3/2) = asecθ∫ x/(a + x)^(3/2) dx= ∫ (atanθ)(asecθ)/(a

利用换元法,令x=atanβ,则dx=asecβdβ ∫dx/(x+a)^(3/2)=∫asecβdβ/asecβ=(1/a)∫dβ/secβ=(1/a)∫cosβdβ=(1/a)sinβ + c=(1/a)tanβcosβ + c=(1/a)tanβ/secβ + c=(1/a)tanβ/√(1+tanβ) +c=(1/a)(x/a)/√[(1+(x/a)] + c=x/[a√(x+a) +c (c为常数)

∫1/(x^2+a^2)*(x^2+a^2)dx=[1/(b^2-a^2)]∫1/(x^2+a^2)-1/(x^2+b^2)dx=[1/(b^2-a^2)][(1/a)arctan(x/a)-(1/b)arctan(x/b)]+c ∫1/(x^2+a^2)dx=(1/a)arctan(x/a)+c

∫x^3/(x^2+a^2)^3/2dx=1/2∫x^2/(x^2+a^2)^3/2d(x^2+a^2)=1/2【∫1/(x^2+a^2)^1/2d(x^2+a^2)-∫a^2/(x^2+a^2)^3/2d(x^2+a^2)】=1/2【2(x^2+a^2)^1/2+2a^2/(x^2+a^2)^1/2】+C=(x^2+a^2)^1/2+a^2/(x^2+a^2)^1/2+C

令x=atanu所以tanu=x/asinu=x/√x^2+a^2cosu=a/√x^2+a^2sin2u= 2ax/x^2+a^2dx=a(secu)^2du原积分=∫a^2[(tanu)^2-1]a(secu)^2du /[a^4(secu)^4]=(1/a)∫[(sinu)^2-(cosu)^2]du=-(1/a)∫cos2udu= -sin2u/2a+C= -x/(x^2+a^2)+C

令x=atanu,则dx=a(secu)^2 du∫(x^2+a^2)^(3/2) dx=∫ (a^4)(secu)^3(secu)^2 du=(a^4)∫ (secu)^5 du=(a^4)[1/4tanu(secu)^3+3/4∫(secu)^3 du]=(a^4)[1/4tanu(secu)^3+

∫x / (x+a)^2/3 dx= 1/2 ∫ 1 / (x+a)^2/3 dx ------ dx = 2x dx = 1/2 ∫(x+a)^(- 2/3) d(x+a) --------d(x+a) = dx = 3/2 * (x+a)^(1/3) + C -------- dx^n = x^(n + 1) / (n + 1)

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