2个回答
展开全部
换元t=x^(1/6),dx=dt^6=6t^5dt
=∫t³/(t²+1)*6t^5dt
=6∫(t^8-1+1)/(t²+1)dt
=6∫(t^4-1)(t²-1)+1/(t²+1)dt
=6t^7/7-6t^5/5-2t³+6t+6arctant+C
=∫t³/(t²+1)*6t^5dt
=6∫(t^8-1+1)/(t²+1)dt
=6∫(t^4-1)(t²-1)+1/(t²+1)dt
=6t^7/7-6t^5/5-2t³+6t+6arctant+C
追答
换元t=√(1-1/(x+1)),则x=1/(1-t²)-1,dx=2t/(1-t²)²dt,
=∫t(1-t²)*2t/(1-t²)²dt
=2∫t²/(1-t²)dt
=∫1/(1+t)+1/(1-t)-2dt
=ln|1+t|-ln|1-t|-2t+C
追问
前一题中t^4-1应为+号。而且最后结果要换回x。不过还是谢谢啦
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