Question

不定积分又一题帮我看看这道http://zhidao.baidu.com/question/189694153.html

Answer 1

原式=∫(x^4+x^2-x^2)/(1+x^2)dx

=∫[x^2-x^2/(1+x^2)]dx

=∫[x^2-(x^2+1-1)/(1+x^2)]dx

=∫[x^2-1+1/(1+x^2)]dx

=(1/3)x^3-x+arctanx+C