Question

求不定积分 1/x^2·根号下x^2+4 x/sinx^2

Answer 1

∫ √(x^2+4)/x^2 dx

let

x=2tany

dx = 2(secy)^2 dy

∫ √(x^2+4)/x^2 dx

=(1/2)∫ [secy/(tany)^2] dy

=(1/2)∫ [cosy/(siny)^2] dy

= -(1/2)(1/siny) + C

=-(1/2)[√(x^2+4) /x] + C

∫ x/sin(x^2) dx

=(1/2)∫ csc(x^2) dx^2

=(1/2)ln|csc(x^2)-cot(x^2)| + C