计算下列不定积分
展开全部
设:t=e^(2x)
则:x=(lnt)/2,dx=1/(2t)dt
∫dx/[1+e^(2x)]
= (1/2)∫1/[t(1+t)]dt
= (1/2)∫[(1+t)-t]/[t(1+t)]dt
= (1/2)∫[1/t - 1/(1+t)]dt
= (1/2)[ln|t|-ln|1+t|]+C
= (1/2)[ln|e^(2x)| - ln|1+e^(2x)]+C
= x-(1/2)ln|1+e^(2x)|+C
则:x=(lnt)/2,dx=1/(2t)dt
∫dx/[1+e^(2x)]
= (1/2)∫1/[t(1+t)]dt
= (1/2)∫[(1+t)-t]/[t(1+t)]dt
= (1/2)∫[1/t - 1/(1+t)]dt
= (1/2)[ln|t|-ln|1+t|]+C
= (1/2)[ln|e^(2x)| - ln|1+e^(2x)]+C
= x-(1/2)ln|1+e^(2x)|+C
追问
第二步的dx不懂
追答
∫dx/[1+e^(2x)]
= (1/2)∫1/[t(1+t)]dt 【dx换成1/(2t)dt,二分之一提到积分外】
= (1/2)∫[(1+t)-t]/[t(1+t)]dt 【分子+t再-t,值不变】
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