# Inverse function of $f(x)=e^x+x-1$.

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.

• Will Lambert W function will play any role here – Daniel Jul 3 at 5:38
• It does not have an explicit inverse as such. – copper.hat Jul 3 at 5:40

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*}

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 ^_^