求积分: ∫√(1+4x^2)dx求详细过程

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求积分: ∫√(1+4x^2)dx求详细过程

求积分: ∫√(1+4x^2)dx求详细过程
求积分: ∫√(1+4x^2)dx
求详细过程

求积分: ∫√(1+4x^2)dx求详细过程
令x=(1/2)tanu,则:
cosu=√{(cosu)^2/[(cosu)^2+(sinu)^2]}=√{1/[1+(tanu)^2]}
=1/√(1+4x^2),
√(1+4x^2)=√[1+(tanu)^2]=1/cosu,dx=(1/2)[1/(cosu)^2]du.
∴原式=(1/2)∫(1/cosu)[1/(cosu)^2]du
   =(1/2)∫[1/(cosu)^3]du
   =(1/2)∫[cosu/(cosu)^4]du
   =(1/2)∫{1/[1-(sinu)^2]^2}d(sinu)
再令sinu=t,则:
原式=(1/2)∫[1/(1-t^2)^2]dt
  =(1/8)∫[(1+t+1-t)/(1+t)(1-t)]^2dt
  =(1/8)∫[1/(1-t)-1/(1+t)]^2dt
  =(1/8)∫[1/(1-t)^2]dt-(1/4)∫[1/(1+t)(1-t)]dt
   +(1/8)∫[1/(1+t)^2]dt
  =-(1/8)[1/(1-t)^2]d(1-t)-(1/8)∫[(1+t+1-t)/(1+t)(1-t)]dt
   +(1/8)∫[1/(1+t)^2]d(1+t)
  =(1/8)[1/(1-t)]-(1/8)∫[1/(1-t)]dt-(1/8)∫[1/(1+t)]dt
   -(1/8)[1/(1+t)]
  =1/(8-8t)-1/(8+8t)-(1/8)ln|1-t|-(1/8)ln|1+t|+C
  =1/(8-8sinu)-1/(8+8sinu)-(1/8)ln(1-t^2)+C
  =16sinu/[64-64(sinu)^2]-(1/8)ln[1-(sinu)^2]+C
  =sinu/[4(cosu)^2]-(1/4)ln|cosu|+C
  =tanu/(4cosu)-(1/4)ln|cosu|+C
  =2x/{4[1/√(1+4x^2)]}-(1/4)ln|1/√(1+4x^2)|+C
  =(1/2)x√(1+4x^2)+(1/8)ln(1+4x^2)+C

∫√(1+4x^2)dx
=X+(4/3)x³+C