If #f(x)=1/(x-1)# how do you solve #intd(f^-1(x))# =?

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#f(x)=1/(x-1)# => #intd(f^-1(x))#=?

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Answer 1 · 2018-04-08T13:07:21

#intd(f^-1(x))=f^-1(x)=(x+1)/x#

We have #f(x)=1/(x-1)# and want to find #intd(f^-1(x))#

We start by finding #f^-1(x)#

#f(x)=1/(x-1)#

#rArrxf(x)-f(x)=1#

#rArrf(x)+1=xf(x)#

#rArrx=(f(x)+1)/(f(x))#

#rArrf^-1(x)=(x+1)/x#

Now #intd(f^-1(x))# is the integral of the derivative #df^-1"/"dx#, ie., it is just #f^-1(x)#

So #intd(f^-1(x))=f^-1(x)=(x+1)/x#