Question

求不定积分，要有过程

Answer 1

let
x=atanu
dx=a(secu)^2 du
∫ x^3/(x^2+a^2)^(3/2) dx
=∫ [(atanu)^3/(asecu)^3 ] . [a(secu)^2 du]
=a∫ (tanu)^3/secu du
=a∫ (sinu)^3/(cosu)^2 du
=-a∫ (sinu)^2/(cosu)^2 dcosu
=a∫ [(cosu)^2-1 ]/(cosu)^2 dcosu
=a∫ [ 1- 1/(cosu)^2 ] dcosu
=a[ cosu + 1/cosu ] + C
=a[ a/√(x^2+a^2) + √(x^2+a^2)/a ] +C
=a^2/√(x^2+a^2) + √(x^2+a^2) + C