Question

能帮忙解一下这些题目吗？要过程，谢谢

Answer 1

(1)
令√(x+1)=t
x为(0→3)
t为(1→2)
x+1=t²
x=t²-1
dx=2tdt
原式=∫(t²-1)t*2tdt
=2∫(t^4-t²)dt
=2(t^5/5-t³/3)+c
=2(32/5-8/3-1/5+1/3)
=2(31/5-7/3)
再通分下即可。
(2)
令√(x+1)=t
x为(0→1)
t为(1→√2)
x+1=t²
x=t²-1
dx=2tdt
原式=∫(t²-1)*2tdt/t
=2∫(t²-1)dt
=2(t³/3-t)+c
=2((√2)³-√2-1/3+1)
再通分下即可。
(3)
令√x=t
x为(9→16)
t为(3→4)
x=t²
dx=2tdt
原式=∫2tdt/(t²-2t)
=2∫dt/(t-2)
=2ln|t-2|+C
=2(ln2-ln1)
=2ln2
(4)
令√x=t
x为(1→9)
t为(1→3)
x=t²
dx=2tdt
原式=∫2tdt/(t²+t)
=2∫dt/(t+1)
=2ln|t+1|+C
=2(ln4-ln2)
=2ln2
(5)
令√x=t
x为(0→4)
t为(0→2)
x=t²
dx=2tdt
原式=∫2tdt/(1+t)
=2∫(t+1-1)dt/(t+1)
=2[∫dt-∫dt/(t+1)]
=2(t-ln|t+1|)+C
=2(2-ln3-0)
=4-2ln3
(6)
令√(x+1)=t
x为(0→3)
t为(1→2)
x+1=t²
x=t²-1
dx=2tdt
原式=∫(t²-1)*2tdt/(1+t)
=2∫(t+1)(t-1)*tdt/(1+t)
=2∫(t-1)*tdt
=2∫(t²-t)dt
=2(t³/3-t²/2)+C
=2(8/3-4/2-1/3+1/2)
=2(7/3-3/2)
=14/3-3
=5/3

追答

写得累死了。。。

追问

谢谢🙏