$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Ei{\operatorname{Ei}}$
\begin{align}
\int_0^1\!-\Wm(-x\,\exp(-x))\,dx
&=\tfrac16\pi^2+\tfrac12
\tag{1}\label{1}
,\\
\int_0^1\!-\Wp(-\tfrac1x\,\exp(-\tfrac1x))\,dx
&=1-\gamma
\tag{2}\label{2}
,\\
\int_0^1\!-\left( \Wm (-x\,\exp(-x)) \right)^{-1}\,dx
&=\gamma
\tag{3}\label{3}
,\\
\int_0^1\!-\Wp(-\tfrac1x\,\exp(-\tfrac1x)
-\left( \Wm (-x\,\exp(-x)) \right)^{-1}\,dx
&=1
\tag{4}\label{4}
,\\
\int_0^1\!\frac{\Wp(-\tfrac1x\,\exp(-\tfrac1x))}
{ \Wm (-\tfrac1x\,\exp(-\tfrac1x))}\,dx
&=1-\ln2
\tag{5}\label{5}
.
\end{align}
\begin{align}
\int_0^1 \Wp(-\tfrac t\e)\,\ln t \, dt
&=
5-\e\,(1+\gamma+\Ei(1,1))
\tag{6}\label{6}
.
\end{align}
\begin{align}
\int_0^1
\left(\Big(-\Wp(-\tfrac t\e)\Big)^{-\tfrac1\e}
-\Big(-\Wm(-\tfrac t\e)\Big)^{-\tfrac1\e}\right)
\, dt
&=
-\tfrac1\e\Gamma(-\tfrac1\e)
\tag{7}\label{7}
,\\
\int_0^1
\left(\Big(-\Wm(-\tfrac t\e)\Big)^{\tfrac1\e}
-\Big(-\Wp(-\tfrac t\e)\Big)^{\tfrac1\e}\right)
\, dt
&=
\tfrac1\e\Gamma(\tfrac1\e)
\tag{8}\label{8}
.
\end{align}
\begin{align}
\int_0^1
\left(\sqrt{-\Wm(-\tfrac t\e)}-\sqrt{-\Wp(-\tfrac t\e)}\right)
\, dt
&=
\frac{\e\sqrt\pi}4
\tag{9}\label{9}
,\\
\int_0^1 \left(\frac 1{\sqrt{-\Wp(-\tfrac t\e)}}-\frac 1{\sqrt{-\Wm(-\tfrac t\e)}}\right) \,dt
&=
\frac{\e\sqrt\pi}2
\tag{10}\label{10}
.
\end{align}
And this one is also one of
integrals-invariant-to-the-choice-of-the-real-branch-of-the-labmert-W-function:
\begin{align}
\int_0^1 \left(2\,\sqrt{-\W(-\tfrac t\e)}+\frac 1{\sqrt{-\W(-\tfrac t\e)}} \right)\, dt
&=\int_0^1 \left(2\,\sqrt{-\Wp(-\tfrac t\e)}+\frac 1{\sqrt{-\Wp(-\tfrac t\e)}} \right)\, dt
\\
&=\int_0^1 \left(2\,\sqrt{-\Wm(-\tfrac t\e)}+\frac 1{\sqrt{-\Wm(-\tfrac t\e)}} \right)\, dt
\\
&=4
\tag{11}\label{11}
.
\end{align}
$\endgroup$
Note that $\sin(t)^n$ can be written as a linear combination of $\sin(k t)$ (if $n$ is odd) or $\cos(k t)$ (if $n$ is even), and you can get antiderivatives for $\sin(k W(x))$ and $\cos(k W(x))$.
$$\int \!\sin \left( k{\rm W} \left(x\right) \right) \,{\rm d}x= \left(
{\frac {x}{{k}^{2}+1}}+2\,{\frac {{k}^{2}x}{ \left( {k}^{2}+1 \right)
^{2}{\rm W} \left(x\right)}} \right) \sin \left( k{\rm W} \left(x
\right) \right) + \left( -{\frac {kx}{{k}^{2}+1}}-{\frac {k \left( {k}
^{2}-1 \right) x}{ \left( {k}^{2}+1 \right) ^{2}{\rm W} \left(x\right)
}} \right) \cos \left( k{\rm W} \left(x\right) \right)
$$
$$ \int \!\cos \left( k{\rm W} \left(x\right) \right) \,{\rm d}x= \left(
{\frac {kx}{{k}^{2}+1}}+{\frac {k \left( {k}^{2}-1 \right) x}{ \left(
{k}^{2}+1 \right) ^{2}{\rm W} \left(x\right)}} \right) \sin \left( k
{\rm W} \left(x\right) \right) + \left( {\frac {x}{{k}^{2}+1}}+2\,{
\frac {{k}^{2}x}{ \left( {k}^{2}+1 \right) ^{2}{\rm W} \left(x\right)}
} \right) \cos \left( k{\rm W} \left(x\right) \right)
$$
Alternatively, express in terms of complex exponentials:
$$ \int \exp(ik {\rm W}(x))\; dx = {\frac {{{\rm e}^{ \left( ik+1 \right) {\rm W} \left(x\right)}}
\left( (ik+1) {\rm W} \left(x\right)+ik \right) }{
\left( ik+1 \right) ^{2}}}
$$
Best Answer
For $t \geqslant 1,$ because the function $t \mapsto 1/t$ is convex, $$ \ln t \leqslant \frac{t - 1}2\left(1 + \frac1t\right) = \frac{t^2 - 1}{2t}, $$ therefore $$ t\ln t \leqslant \frac{t^2 - 1}2 = (t - 1) + \frac{(t - 1)^2}2 \leqslant e^{t - 1} - 1 \quad (t \geqslant 1). $$ Equivalently, because the function $t \mapsto e^{t - 1} - 1$ is strictly increasing on $[1, \infty),$ $$ (\ln(x + 1) + 1)\ln(\ln(x + 1) + 1) \leqslant x \quad (x \geqslant 0). $$ Therefore $\ldots$