Question

求下列定积分的过程

Answer 1

望采纳，谢谢啦。

Answer 2

（1）令t=1+x^(1/3)，则x=(t-1)^3，dx=3(t-1)^2dt
原式=∫(1,2) 3(t-1)^2/tdt
=3∫(1,2) (t-2+1/t)dt
=3*(t^2/2-2t+ln|t|)|(1,2)
=3*(2-4+ln2-1/2+2)
=3ln2-3/2
（2）令t=√x-1，则x=(t+1)^2，dx=2(t+1)dt
原式=∫(1,2) (t+1)/t*2(t+1)dt
=2∫(1,2) (t+2+1/t)dt
=(t^2+4t+2ln|t|)|(1,2)
=4+8+2ln2-1-4
=7+2ln2
（3）令t=√(5-4x)，则x=(5-t^2)/4，dx=-t/2dt
原式=∫(3,1) 1/t*(-t/2)dt
=(1/2)*∫(1,3)dt
=(t/2)|(1,3)
=3/2-1/2
=1
（4）令t=√(x+1)，则x=t^2-1，dx=2tdt
原式=∫(1,2) (t^2-1)t*2tdt
=2∫(1,2) (t^4-t^2)dt
=[(2/5)*t^5-(2/3)*t^3]|(1,2)
=64/5-16/3-2/5+2/3
=62/5-14/3
=116/15