如题所述
1、令x=t^6,则dx=6t^5dt
√x=t³,³√x=t²
原式=∫6t^5dt/t³(1+t²)
=6∫t²dt/(1+t²)
=6∫(1+t²-1)dt/(1+t²)
=6∫dt-6∫dt/(1+t²)
=6t-6arctant+C
=6x^(1/6)-6arctan(x)^(1/6)+C
2、原式=∫(0,1)xd(e^x)
=xe^x|(0,1)-∫(0,1)e^xdx
=xe^x|(0,1)-e^x|(0,1)
=(e-0)-(e-1)
=1