计算下列定积分

Question

5.7.9.10

Answer 1

设根号(e^x-1) =t
t^2 +1=e^x
x=ln(t^2 +1)
代入得
∫t dln(t^2 +1)
=∫2t^2/(t^2 +1) dt
=2*∫t^2/(t^2 +1) dt
=2*∫(t^2 +1-1)/(t^2 +1) dt
=2*∫[1 -1/(t^2 +1)] dt
=2*[∫1 dt -∫1/(t^2 +1) dt
=2*(t -arctant) +C(常数)
=2*【(e^x-1) -arctan(e^x-1)】+C
=2*【e^x -arctan(e^x-1)】+C(常数都归纳到C)把数字带进去就好追答

当-2＜x＜0时
∫[(x+|x|)/(2+x^2)]dx=∫[(x-x)/(2+x^2)]dx=0
当0≤x＜2时
∫[(x+|x|)/(2+x^2)]dx=∫[(x+x)/(2+x^2)]dx=∫[2x/(2+x^2)]dx=∫[1/(2+x^2)]d(2+x^2)=ln(2+x^2)=ln6-ln2=ln3
所以在-2到2区间上
∫[(x+|x|)/(2+x^2)]dx=0+ln3=ln3

9.∫(2x-x^2)^1/2dx
=∫(1-(x-1)^2)^1/2d(x-1)
令x-1=cosy,则：
=∫siny d（cosy）
=∫（siny）^2dy
=1/2(y-sinycosy)
x-1∈[-1,0],y∈[π/2,π]
代入之后得到：
1/2[(π-π/2)]-1/2[sinπcosπ-sin(π/2)cos(π/2)]
=π/4

11.求定积分(0,a) ∫x²√(a²-x²) dx
原式=(0,a)∫(ax²√[1-(x/a)²]dx
令x/a=sint,则dx=acostdt,x=0时,t=0；x=a时,t=π/2.
故原式=(0,π/2)a⁴∫sin²tcos²tdt=(0,π/2)(a⁴/4)∫sin²(2t)dt=(0,π/2)(a⁴/8)∫sin²(2t)d(2t)
=(0,π/2)(a⁴/16)∫[(1-cos4t)/2]d(4t)=(0,π/2)(a⁴/32)∫[(1-cos4t)d(4t)
=[(a⁴/32)(4t-sin4t)](0,π/2)=(a⁴/32)×(2π)=πa⁴/16

追问

好的，蟹蟹～

追答

嗯