Question

求教一道定积分？

谢谢

Answer 1

更改为π/4,最后答案得π/4+ln2

Answer 2

Answer 3

let
1/[x(1+x^2)]≡ A/x + (Bx+C)/(1+x^2)
=>
1≡ A(1+x^2) + (Bx+C)x
x=0, => A=1
coef. of x^2
A+B= 0
1+B=0
B=-1
coef. of x , => C=0
1/[x(1+x^2)]
≡ A/x + (Bx+C)/(1+x^2)
≡ 1/x - x/(1+x^2)

∫(1->+∞) arctanx/x^2 dx
=-∫(1->+∞) arctanx d(1/x)
=- [arctanx/x]|(1->+∞) + ∫(1->+∞) dx/[x(1+x^2)] dx
=π/4 + ∫(1->+∞) [1/x - x/(1+x^2) ] dx
=π/4 + [ ln|x/√(1+x^2)| ]|(1->+∞)
=π/4 - ln(1/√2)
=π/4 +(1/2) ln2

Answer 4

Answer 5

分部积分