= Solution
Fix $t\in\mathbb R$. The <mean value theorem> for the exponential gives
$$
|e^{tx}-1|\leq |t|x e^{|t|},\qquad0<x<1.
$$
Consequently
$$
|u(x,t)|\leq\frac{|t|e^{|t|}}{\sqrt{1-x}},\qquad
\int_0^1\frac{|t|e^{|t|}}{\sqrt{1-x}}\,dx=2|t|e^{|t|}<\infty.
$$
The function is continuous on $(0,1)$ and hence measurable, so \b[$u(\cdot,t)$ is <Lebesgue integrable> for every real $t$]. In particular, the numerator cancels the apparent $1/x$ singularity at zero: $u(x,t)\to t$ as $x\downarrow0$. Only the integrable square-root singularity at one remains. At $t=0$ the function is identically zero.
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