# problem with integration with backwards substitution - finding inverse: I have: u=g(x)=(e^x -1)^1/2, so x=h(u) = ln(u^2 +1) I can’t work out how to get from one to other. Please show me!

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You should remember that `f'(x)*(f^(-1)(x))' = 1` , hence, you need to find f'(x) such that:

`f'(x) = (1/2)(e^x - 1)^(1/2 - 1)*(e^x - 1)'`

`f'(x) = (e^x)/(2sqrt(e^x - 1))`

Hence `(f^(-1)(x))' = 1/f'(x) = 1/((e^x)/(2sqrt(e^x - 1))) => (f^(-1)(x))' = (2sqrt(e^x - 1))/e^x` .

You may evaluate `f^(-1)(x)` integrating `(f^(-1)(x))'` such that:

`f^(-1)(x) = int (2sqrt(e^x - 1))/e^x dx`

You should come up with the substitution `e^x = u => e^x dx = du`

`int (2sqrt(e^x - 1))/e^x dx = 2int sqrt(u-1)/(u^2)du`

You should come up with the substitution `sqrt(u-1)=t => 1/(2sqrt(u-1)) = dt`

`2 int sqrt(u-1)/(u^2)du = 4 int (t^2/(t^2+1)^2)dt`

Using partial fraction decomposition yields `t^2/(t^2+1)^2 = 1/(t^2+1) - 1/((t^2+1)^2)`

`4 int (t^2/(t^2+1)^2)dt = 4 (int 1/(t^2+1)dt - int 1/((t^2+1)^2)dt)`

`2 int sqrt(u-1)/(u^2)du = 2tan^(-1)sqrt(u-1) - (2sqrt(u-1))/u + c`

`f^(-1)(x) = 2tan^(-1)sqrt(e^x-1) - 2e^(-x)sqrt(e^x - 1) + c`

**Hence, evaluating the inverse of function using integration yields `f^(-1)(x) = 2tan^(-1)sqrt(e^x-1) - 2e^(-x)sqrt(e^x - 1) + c.` **