Question

怎样求微积分

2、3、4的动力学方程怎样求定积分式，急急急

Answer 1

1/[(a-x)^2.(c-x)] ≡ A/(a-x) +B/(a-x)^2 +C/(c-x)
=>
1 ≡ A(a-x)(c-x) +B(c-x) +C(a-x)^2
x=a => B= 1/(c-a)
x=c =>C = 1/(a-c)^2
coef. of x^2
A+C =0
A=-C = -1/(a-c)^2
=>
1/[(a-x)^2.(c-x)]
≡ -[1/(a-c)^2] [1/(a-x)] +[1/(c-a)] [1/(a-x)^2] +[1/(a-c)^2][1/(c-x)]
dx/dt = k3.(a-x)^2.(c-x)
∫ dx/[(a-x)^2.(c-x)] = ∫ k3. dt
k3t
=∫ { -[1/(a-c)^2] [1/(a-x)] +[1/(c-a)] [1/(a-x)^2] +[1/(a-c)^2][1/(c-x)]} dx
=[1/(a-c)^2] ln|a-x| + [1/(c-a)] [1/(a-x) ] -[1/(a-c)^2]ln|c-x| + C'

Answer 2

分离变量然后积分，左侧分解为部分分式去积分