Question

x^2/（1+x^2）dx求不定积分 基础有点差 步骤详细点

Answer 1

∫x²/（x²+1)dx
=∫ （x²+1-1）/(x²+1)dx
=∫ 1-[1/(x²+1)]dx
=x-arctanx+C
令x = tanu，x²+1=tan²u+1=sec²u，dx = sec²u du
= ∫ tan²u/sec⁴u * (sec²u du)
= ∫ sin²u/cos²u * cos²u du
= ∫ (1 - cos2u)/2 du
= u/2 - (1/4)sin2u + C
= (1/2)arctanx - (1/2) * x/√(1 + x²) * 1/√(1 + x²) + C
= (1/2)arctanx - x/[2(1 + x²)] + C

追问

看图片里的，标题上分母少了个平方

追答

令x = tanu，x²+1=tan²u+1=sec²u，dx = sec²u du
= ∫ tan²u/sec⁴u * (sec²u du)
= ∫ sin²u/cos²u * cos²u du
= ∫ (1 - cos2u)/2 du
= u/2 - (1/4)sin2u + C
= (1/2)arctanx - (1/2) * x/√(1 + x²) * 1/√(1 + x²) + C
= (1/2)arctanx - x/[2(1 + x²)] + C