求反常积分，题目如图。 我求解了一次结果是1⼀4ln3+π⼀4 参考答案是π⼀4.

let
1/[(1+x)(1+x^2)] ≡ A/(1+x) +(Bx+C)/(1+x^2)
=>
1≡ A(1+x^2) +(Bx+C)/(1+x)
x=-1 => A =1/2
coef. of x^2
A+B=0
B=-1/2
coef. of constant
A+C =1
C=1/2
1/[(1+x)(1+x^2)] ≡ (1/2) [1/(1+x) -(x-1)/(1+x^2)]
∫(0->+∞) dx/[(1+x)(1+x^2)]
=(1/2) ∫(0->+∞) [1/(1+x) -(x-1)/(1+x^2)] dx
=(1/2) ∫(0->+∞) [1/(1+x) -x/(1+x^2) + 1/(1+x^2) ] dx
=(1/2) [ ln|1+x| - (1/2)ln|1+x^2| + arctanx ]|(0->+∞)
=(1/2) lim(x->+∞) [ ln|1+x| - (1/2)ln|1+x^2| + arctanx ]
=(1/2) [ lim(x->+∞) ln|1+x| - (1/2)ln|1+x^2| ] + π/4
=(1/2)[ lim(x->+∞) ln|(1+x)/√(1+x^2) ] + π/4
=(1/2)(0) +π/4
=π/4