1\(x^8*(1+x^2))的不定积分是多少

Question

1\(x^8*(1+x^2))的不定积分是多少

Answer 1

用倒代换.令x=1/t,则dx=-dt/t²,原积分=-∫t^8/(t^2+1)dt=-∫t^6dt+∫t^6/(1+t^2)dt=-∫t^6dt+∫t^4dt-∫t^4/(1+t^2)dt=-∫t^6dt+∫t^4dt-∫t^2dt+∫t^2/(1+t^2)dt=-∫t^6dt+∫t^4dt-∫t^2dt+∫dt-∫dt/（1+t²）=-1/7t^7+1/5t^5-1/3t^3+t-arctant+C=-1/(7x^7)+1/(5x^5)-1/(3x^3)+1/x-arctan(1/x)+C