求一积分设f(x^2
设f(x^2-1)=lnx^2/(x^2-2),且f[g(x)]=lnx,求∫g(x)dx
f(x?-1)=ln[x?/(x?-2)]=ln[(x?-1+1)/(x?-1-1)] 故f(x)=ln[(x+1)/(x-1)] 则f[g(x)]=ln[g(x)+1/g(x)-1]=lnx 即[g(x)+1]/[g(x)-1]=x 解得g(x)=(x+1)/(x-1) 故∫g(x)dx=∫(x-1+2)/(x-1)dx=x+2ln|x-1|+C