Question

求dx/√x(1+3次根号下x)的不定积分…详细过程

Answer 1

设（6次根号X）=t,则原式可变换为
∫1/[t^3(1+t^2)]dt^6
= 6∫t^2/(1+t^2)dt
= 6∫(1+t^2-1)/(1+t^2)dt
=6∫dt-6∫1/(1+t^2)dt
=6t-arctant+C
再把原来的假设代回原式就行啦