Question

数学题，哪位大神火速解答一下，如题目有问题请指正？

求大神解答

Answer 1

(2)
y= [(x+1))x+2)/(x+3)]^(1/3)
lny =(1/3)[ ln(x+1)+ln(x+2)-3ln(x+3)]
y'/y=(1/3)[ 1/(x+1)+1/(x+2)-3/(x+3)]
y'=(1/3)[ 1/(x+1)+1/(x+2)-3/(x+3)] .[(x+1))x+2)/(x+3)]^(1/3)
(3)
∫ (1/x)(lnx)^2 dx
=∫ (lnx)^2 dlnx
=(1/3)(lnx)^3 +C
(4)
∫ lnx dx
=xlnx -∫dx
=xlnx -x + C
4.
(1)
let
u=π/2-x
du =-dx
x=0, u=π/2
x=π/2, u=0
∫(0->π/2) f(sinx)dx
=∫(π/2->0) f(cosu) (-du)
=∫(0->π/2) f(cosu) du
=∫(0->π/2) f(cosx) dx
(2)
∫(0->1-) dx/√(1-x^2)
=[arcsinx](0->1-)
=π/2
(3)
4y''-20y'+25y=0
The aux.equation
4r^2-20r +25=0
(2r-5)^2=0
r=5/2
通解
y=(A+Bx)e^[(5/2)x]

Answer 2

详见下图，图片可点击放大：