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x^2/√1+4x^2积分怎么求?高数题

换元法,很简单而已,令x = 1/2 tanz,dx = 1/2 seczdz√(1 + 4x) = secz∫ x√(1 + 4x) dx= ∫ (1/8)tanz secz (1/2)secz dz= (1/16)∫ tanzsecz d(secz)= (1/16)∫ (secz - 1)secz d(secz)= (1/16)∫ (secz - secz) d(secz)= (1/

原式=1/2∫√(1+4ρ^2)*ρ^2dρ^2 (令ρ^2=t)=1/2∫√(1+4t)*tdt然后积分,用分步

答案略长,有问题就问,没问题就采纳吧

∫x √(1+4x^2) dx=1/8∫ √(1+4x^2) d(1+4x^2)=1/12 (1+4x^2)^3/2

令x=tant,积分化为∫sectdt/(tant)^2 =∫costdt/(sint)^2 =-1/sint+C=-√((x^2+1)]/x+C

令 2x = tanu,则 dx = (secu)^2 / 2 du,原式 = ∫secu/2 du= 1/2 ln|tanu+secu| + C = 1/2 ln|2x+√(1+4x^2)| + C.

∫1/(x^2-4x+2)=∫1/(x^2-4x+4-2)=∫1/[(x-2)^2-(根号2)^2]=∫1/[(x-2+根号2)(x-2-根号2)]=(-1/2根号2)[∫1/(x-2+根号2)-∫1/(x-2-根号2)]=(-1/2根号2)[ln(x-2+根号2)-ln(x-2-根号2)]=(-1/2根号2)[ln(x-2+根号2)/(x-2-根号2)]

答:∫ x√(1+4x^2) dx=(1/2)*∫ √(1+4x^2) d(x^2)=(1/8)*∫ √(1+4x^2) d(1+4x^2)=(1/8)*(2/3)*(1+4x^2)^(3/2)+C=(1/12)*(1+4x^2)*√(1+4x^2)+C

=§x(1+x^2)^1/2/x^2dx=1/2§(x^2+1)^1/2/x^2dx^2换元即求§(t+1)^1/2/tdt再另(t+1)^1/2=y剩下就是有理函数积分,易得结果!

∫(x-2)*√(x^2+4x+1)dx=1/3 (-11 - 2 x + x^2) Sqrt[1 + 4 x + x^2] + 6 Log[2 + x + Sqrt[1 + 4 x + x^2]]+c

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