Answer
$$\sqrt{2}+\ln (1+\sqrt{2}) $$
Work Step by Step
Since
$$\mathbf{r}=(\sqrt{2} t) \mathbf{i}+(\sqrt{2} t) \mathbf{j}+\left(1-t^{2}\right) \mathbf{k} $$
Then
\begin{align*}
\mathbf{v}&=\sqrt{2} \mathbf{i}+\sqrt{2} \mathbf{j}-2 t \mathbf{k} \\|\mathbf{v}|&=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}+(-2 t)^{2}}\\
&=\sqrt{4^{2}+4 t}=2 \sqrt{1+t^{2}}
\end{align*}
Hence
\begin{align*}
\text{Length}&=\int_{0}^{1} 2 \sqrt{1+t^{2}} d t\\
&=\left[2\left(\frac{t}{2} \sqrt{1+t^{2}}+\frac{1}{2} \ln (t+\sqrt{1+t^{2}})\right)\right]_{0}^{1}\\
&=\sqrt{2}+\ln (1+\sqrt{2})
\end{align*}
