Question

大学数学问题，求不定积分，蟹蟹

Answer 1

Answer 2

let
x^3= √3tanu
3x^2 dx =√3(secu)^2 du
∫ dx/[x(x^6+3)]
=(1/3)∫ 3x^2 dx/[x^3.(x^6+3)]
=(1/3)∫ √3(secu)^2 du/[3√3. tanu.(secu)^2]
=(1/9)∫ du/tanu
=(1/9)ln|sinu | + C
=(1/9)ln| x^3/(x^6+3) | + C