Question

高数不定积分，求详细解答

Answer 1

Answer 2

{" integral 1/(x^2+1)^2 dx = 1/2 (x/(x^2+1)+tan^(-1)(x))+constant", \
"Take the integral:
integral 1/(x^2+1)^2 dx
For the integrand 1/(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) du
Write cos^2(u) as 1/2 cos(2 u)+1/2:
= integral (1/2 cos(2 u)+1/2) du
Integrate the sum term by term and factor out constants:
= 1/2 integral cos(2 u) du+1/2 integral 1 du
For the integrand cos(2 u), substitute s = 2 u and ds = 2 du:
= 1/4 integral cos(s) ds+1/2 integral 1 du
The integral of cos(s) is sin(s):
= (sin(s))/4+1/2 integral 1 du
The integral of 1 is u:
= (sin(s))/4+u/2+constant
Substitute back for s = 2 u:
= u/2+1/4 sin(2 u)+constant
Substitute back for u = tan^(-1)(x):
= (x^2 tan^(-1)(x)+x+tan^(-1)(x))/(2 x^2+2)+constant
Which is equal to:
Answer: |
| = 1/2 (x/(x^2+1)+tan^(-1)(x))+constant"}

Answer 3

直接分部积分。

追问

写一下过程可以吗

追答

左边∫dx/(1+x²)=x/(1+x²)+2∫【x²/(1+x²)²】dx
在第二项分子上+1-1拆成两个积分，得到
=x/(1+x²)+2∫dx/(1+x²)-2∫dx/(1+x²)²右边，
从中解出所求积分=(1/2)【arctanx+（x/(1+x²)）】+C。

Answer 4