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Please find the inverse function of $f(x)=e^x+x-1$. I want to integrate the inverse functions of this but I am not able to find out any possible way to do so.

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  • $\begingroup$ Will Lambert W function will play any role here $\endgroup$ – Daniel Jul 3 at 5:38
  • $\begingroup$ It does not have an explicit inverse as such. $\endgroup$ – copper.hat Jul 3 at 5:40
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You can solve $y = e^x + x - 1$ for $x$ with the Lambert W function, as you suspected: \begin{align*} y = e^x + x - 1 &\implies y -x + 1 = e^x \\ &\implies (y - x + 1)e^{-x} = 1 \\ &\implies (y - x + 1)e^{y - x + 1} = e^{y + 1} \\ &\implies y - x + 1 = W(e^{y + 1}) \\ &\implies x = y + 1 - W(e^{y + 1}). \end{align*}

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Don't forget there is a formula for integrating inverse functions! Depending on what you want the bounds of integration to be, you might not need to invert the function at all.

If $c = e^a + a - 1$ and $d = e^b + b - 1$, for any two reals $a$ and $b$, then

$$\int_c^d (e^x + x - 1)^{-1} dx + \int_a^b (e^x + x - 1) = bd - ac$$

More explicitly, for any $a$ and $b$ we have

$$ \int_{e^a + a - 1}^{e^b + b - 1} (e^x + x - 1)^{-1} = b(e^b + b -1) - a(e^a + a - 1) - (e^b - e^a + \frac{1}{2}(b^2 - a^2) - (b - a)) $$

There is also a formula (on the same page) for taking an antiderivative, rather than integrating with bounds. However the antiderivative formula is less useful because it requires a closed form for $(e^x + x - 1)^{-1}$, whereas the above formula allows us to compute (certain) integrals without ever looking at a $W$-function!


I hope this helps ^_^

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