Question

x^4/(x^2+1)怎么求不定积分

Answer 1

求不定积分∫[x⁴/(x²+1)]dx
解：∫[x⁴/(x²+1)]dx=∫[x²-1+1/(1+x²)]dx=∫x²dx-∫dx+∫dx/(1+x²)=(x³/3)-x+arctanx+C

Answer 2

解：∫ x^4 / (x^2+1) dx
= ∫ (x^4-1+1) / (x^2+1) dx
= ∫ (x^4-1) / (x^2+1) dx + ∫ 1 / (x^2+1) dx
= ∫ (x^2-1) dx + arctanx
= 1/3 x^3 - x + arctanx + C

追问

非常感谢