Question

不定积分问题求解谢谢了两题

Answer 1

Answer 2

追问

第二题好复杂 我根本想不到

追答

第2题是有点复杂。积累经验吧，也没什么。
若满意，请采纳。

Answer 3

(1)
∫x(arctanx)^2 dx
=(1/2) ∫(arctanx)^2 dx^2
=(1/2)x^2.(arctanx)^2 - ∫[x^2.(arctanx)/(1+x^2)] dx
=(1/2)x^2.(arctanx)^2 - ∫[1 - 1/(1+x^2)] (arctanx) dx
=(1/2)x^2.(arctanx)^2 - ∫arctanx dx +∫ arctanx/(1+x^2) dx
=(1/2)x^2.(arctanx)^2 - xarctanx +∫x/(1+x^2) dx +∫ arctanx darctanx
=(1/2)x^2.(arctanx)^2 - xarctanx +(1/2)ln|1+x^2| + (1/2)(arctanx)^2 + C
(2)
let
(x^2+1) /[(x+1)^2.(x-3)]≡A/(x+1) +B/(x+1)^2 +C/(x-3)
=>
x^2+1≡A(x+1)(x-3) +B(x-3)+C(x+1)^2
x=3, => C = 5/8
x=-1, => B= -1/2
coef. of x^2
A+C =1
A+5/8 =1
A=3/8
(x^2+1) /[(x+1)^2.(x-3)]≡(3/8)[1/(x+1)] -(1/2)[1/(x+1)^2] +(5/8) [1/(x-3)]
∫(x^2+1) /[(x+1)^2.(x-3)] dx
=∫{ (3/8)[1/(x+1)] -(1/2)[1/(x+1)^2] +(5/8) [1/(x-3)] } dx

= (3/8)ln|x+1| +(1/2)[1/(x+1)] +(5/8)ln|x-3| + C