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I want to solve the following integral, where $W$ is the Lambert W function.
\begin{equation} \int \frac{W(e^{4x-3})}{1+W(e^{4x-3})}dx \end{equation} I assume $x \in [0, 1]$.

Can someone please check my solution?

Integrate by substitution with $t = W(e^{4x-3})$.

Then $W^{-1}(t) = te^t=e^{4x-3}$. Thus $t + \log(t) = 4x-3$ and $x = 0.25 (3 + t + \log(t))$.

This gives \begin{equation} dx = 0.25 \left( 1 + \frac{1}{t} \right) dt \end{equation}

So we want to solve \begin{equation} 0.25 \int \frac{t}{1+t} \left( 1 + \frac{1}{t} \right) dt = 0.25 \int 1 dt = 0.25 t + c_0 \end{equation}

Substituting back \begin{equation} \int \frac{W(e^{4x-3})}{1+W(e^{4x-3})}dx = 0.25 W(e^{4x-3}) + c_0 \end{equation}

Is this correct? Is there an easier way to see the integral comes out to this?

neophyte_autodidact
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  • Your method looks fine to me (it's the fastest way I know how to do this problem). Just to be sure, I checked Wolfram Alpha, and it agrees with your solution. – projectilemotion Sep 06 '18 at 17:45
  • Wow, thanks for the Wolfram Alpha link! I didn't know it could do that, but it is so cool! – neophyte_autodidact Sep 06 '18 at 17:52
  • Whenever you have done an indefinite integral, you can check your result by differentiating your answer. – GEdgar Sep 20 '18 at 13:18
  • Yes, I guess I didn't think of that because differentiating functions involving Lambert's $W$ doesn't come naturally to me, but since this is a completely general way of verifying an integral is correct, I probably should have thought of it – neophyte_autodidact Sep 20 '18 at 13:29

1 Answers1

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By using $$\frac{d W(t)}{dt} = \frac{W(t)}{t \, (1 + W(t))}$$ then the integral is even easier to determine. This leads to: \begin{align} I &= \int \frac{W(e^{a x + b})}{1 + W(e^{ax + b})} \, dx \\ &= \frac{1}{a} \, \int \frac{W(e^{ax + b}) \, d(e^{a x + b})}{e^{a x + b} \, (1 + W(e^{ax +b}))} \\ &= \frac{1}{a} \int \frac{d W(e^{a x + b})}{dx} \, dx \\ &= \frac{1}{a} \, W(e^{a x +b}) + c_{0}. \end{align}

Leucippus
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