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?
Best Answer
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}