Question

分解因式法求有限积分

Answer 1

Answer 2

∫(0->1) (2-x^2)/[(1+x)(1+x^2) ]dx
let
(2-x^2)/[(1+x)(1+x^2)] ≡ A/(1+x) + (Bx+C)/(1+x^2)
=>
2-x^2 ≡ A(1+x^2) + (Bx+C)(1+x)
x=-1, => A = 1/2
coef. of x^2
A+B=-1
1/2+B=-1
B=-3/2
coef. of constant
A+C = 2
1/2 + C =2
C= 3/2
∫(0->1) (2-x^2)/[(1+x)(1+x^2) ]dx
=∫(0->1) { (1/2)[1/(1+x)] -(3/2)[ (x-1)/(1+x^2) ] } dx
= (1/2)[ln|1+x|]|(0->1) -(3/4)[ln|1+x^2| ]|(0->1) + (3/2)[ arctanx]|(0->1)
=(1/2)ln2 - (3/4)ln2 + (3/2)(π/4)
=(3/8)π - (1/4)ln2

Answer 3

不知道是不是这样做的