Question

高等数学基础，利用分部积分法求式子

Answer 1

∫ (arctanx)/[x^2.(1+x^2)] dx
=∫ (arctanx)/x^2 dx - ∫ (arctanx)/(1+x^2) dx
=-∫ (arctanx) d(1/x) - (1/2)∫ d(arctanx)^2
=-(arctanx)/x + ∫ dx/[x(1+x^2)] - (1/2)(arctanx)^2
=-(arctanx)/x - (1/2)(arctanx)^2 + ∫ [1/x- x/(1+x^2)]dx
=-(arctanx)/x - (1/2)d(arctanx)^2 + ln|x| - (1/2)ln|1+x^2| + C
consider
1/[x(1+x^2)] ≡ A/x+ (Bx+C)/(1+x^2)
=>
1 ≡ A(1+x^2)+ x(Bx+C)
x=0, => A=1
coef. of x^2
A+B=0
B=-1
coef. of x , => C=0
1/[x(1+x^2)] ≡ 1/x- x/(1+x^2)

Answer 2

先将拆成两项，然后在

追答

接下来，就简单多了。