Question

∫(x+根号1+x^2)^2dx=?

Answer 1

这种不定积分建议先展开后换元
∫[x+√(1+x^2)]^2dx
=∫[2x^2+1+2x*√(1+x^2)]dx
=∫(2x^2+1)dx+2∫[x*√(1+x^2)]dx
对于2∫[x*√(1+x^2)]dx,令x=tant
则2∫[x*√(1+x^2)]dx
=2∫(sect*tant)d(tant)
=2∫(sec^3t*tant)dt
=2∫(sec^2t)d(sect),[注:d(sect)=sect*tant*dt]
=2/3*sec^3t
又x=tant,由辅助三角形知cost=1/√(1+x^2)
故sect=√(1+x^2)
故2∫[x*√(1+x^2)]dx=2/3*(1+x^2)^(3/2)
故∫(2x^2+1)dx+2∫[x*√(1+x^2)]dx
=2/3*x^3+x+2/3*(1+x^2)^(3/2)+C
所以∫[x+√(1+x^2)]^2dx=2/3*x^3+x+2/3*(1+x^2)^(3/2)+C