Question

设f(x^2-1)=lnx^2/(x^2-2),且f[g(x)]=lnx,求∫g(x)dx

Answer 1

设x^2-1=t,则x^2=t+1,

f(t)=ln(t+1)/(t-1)

设(t+1)/(t-1)=x,得t=(x+1)/(x-1)

即g(x)=(x+1)/(x-1)

所以 ∫g(x)dx=x+ln(x-1)