Question

高数求不定积分

Answer 1

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