Question

求这题的不定积分，急用，求过程解答，谢谢！

Answer 1

复制过来的，我认识外文字母啊。

Take the integral:
integral (x^3+1)/(x^2+1)^2 dx
For the integrand (x^3+1)/(x^2+1)^2, use partial fractions:
= integral ((1-x)/(x^2+1)^2+x/(x^2+1)) dx
Integrate the sum term by term:
= integral (1-x)/(x^2+1)^2 dx+ integral x/(x^2+1) dx
For the integrand (1-x)/(x^2+1)^2, substitute x = tan(u) and dx = sec^2(u) du. Then (x^2+1)^2 = (tan^2(u)+1)^2 = sec^4(u) and u = tan^(-1)(x):
= integral cos^2(u) (1-tan(u)) du+ integral x/(x^2+1) dx
Write cos^2(u) as 1-sin^2(u):
= integral (1-sin^2(u)) (1-tan(u)) du+ integral x/(x^2+1) dx
Expanding the integrand (1-sin^2(u)) (1-tan(u)) gives -sin^2(u)-tan(u)+sin^2(u) tan(u)+1:
= integral (-sin^2(u)-tan(u)+sin^2(u) tan(u)+1) du+ integral x/(x^2+1) dx
Integrate the sum term by term and factor out constants:
= integral 1 du- integral sin^2(u) du- integral tan(u) du+ integral sin^2(u) tan(u) du+ integral x/(x^2+1) dx
Write sin^2(u) as 1-cos^2(u):
= integral 1 du- integral sin^2(u) du- integral tan(u) du+ integral (1-cos^2(u)) tan(u) du+ integral x/(x^2+1) dx
Expanding the integrand (1-cos^2(u)) tan(u) gives tan(u)-sin(u) cos(u):
= integral 1 du- integral sin^2(u) du- integral tan(u) du+ integral (tan(u)-sin(u) cos(u)) du+ integral x/(x^2+1) dx
Integrate the sum term by term and factor out constants:
= integral 1 du- integral sin^2(u) du+ integral tan(u) du- integral tan(u) du- integral sin(u) cos(u) du+ integral x/(x^2+1) dx
For the integrand sin(u) cos(u), substitute s = cos(u) and ds = -sin(u) du:
= integral s ds+ integral 1 du- integral sin^2(u) du- integral tan(u) du+ integral tan(u) du+ integral x/(x^2+1) dx
Write sin^2(u) as 1/2-1/2 cos(2 u):
= integral s ds+ integral 1 du- integral (1/2-1/2 cos(2 u)) du- integral tan(u) du+ integral tan(u) du+ integral x/(x^2+1) dx
Integrate the sum term by term and factor out constants:
= integral s ds-1/2 integral 1 du+ integral 1 du+1/2 integral cos(2 u) du- integral tan(u) du+ integral tan(u) du+ integral x/(x^2+1) dx
For the integrand cos(2 u), substitute p = 2 u and dp = 2 du:
= 1/4 integral cos(p) dp+ integral s ds+ integral 1 du-1/2 integral 1 du- integral tan(u) du+ integral tan(u) du+ integral x/(x^2+1) dx
The integral of cos(p) is sin(p):
= (sin(p))/4+ integral s ds+ integral 1 du-1/2 integral 1 du- integral tan(u) du+ integral tan(u) du+ integral x/(x^2+1) dx
The integral of 1 is u:
= (sin(p))/4+ integral s ds-u/2+ integral 1 du- integral tan(u) du+ integral tan(u) du+ integral x/(x^2+1) dx
For the integrand x/(x^2+1), substitute w = x^2+1 and dw = 2 x dx:
= (sin(p))/4+ integral s ds-u/2+ integral 1 du- integral tan(u) du+ integral tan(u) du+1/2 integral 1/w dw
The integral of 1/w is log(w):
= (sin(p))/4+ integral s ds-u/2+ integral 1 du- integral tan(u) du+ integral tan(u) du+(log(w))/2
The integral of s is s^2/2:
= (sin(p))/4+s^2/2-u/2+ integral 1 du- integral tan(u) du+ integral tan(u) du+(log(w))/2
The integral of tan(u) is -log(cos(u)):
= (sin(p))/4+s^2/2-u/2-log(cos(u))+ integral 1 du- integral tan(u) du+(log(w))/2
The integral of tan(u) is -log(cos(u)):
= (sin(p))/4+s^2/2-u/2+ integral 1 du+(log(w))/2
The integral of 1 is u:
= (sin(p))/4+s^2/2+u/2+(log(w))/2+constant
Substitute back for w = x^2+1:
= (sin(p))/4+s^2/2+u/2+1/2 log(x^2+1)+constant
Substitute back for p = 2 u:
= s^2/2+u/2+1/4 sin(2 u)+1/2 log(x^2+1)+constant
Substitute back for s = cos(u):
= u/2+1/4 sin(2 u)+(cos^2(u))/2+1/2 log(x^2+1)+constant
Substitute back for u = tan^(-1)(x):
= ((x^2+1) log(x^2+1)+(x^2+1) tan^(-1)(x)+x+1)/(2 (x^2+1))+constant
Which is equal to:
Answer: |
| = 1/2 ((x+1)/(x^2+1)+log(x^2+1)+tan^(-1)(x))+constant

追问

这么一大串，看花了我的眼。。这题我自己已经做出来了，不过还是要谢谢你，麻烦你了。

追答

ctrl A, Ctrl C,
Ctrl V
我一点都不眼花，嘿嘿。

追问

嘿嘿，谢谢啦。

追答

甭客气