Evaluate
x\sqrt{2dx_{3}}+С
\left(d\geq 0\text{ and }x_{3}\geq 0\right)\text{ or }\left(x_{3}\leq 0\text{ and }d\leq 0\right)
Differentiate w.r.t. x
\sqrt{2dx_{3}}
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\sqrt{2x_{3}d}x
Find the integral of \sqrt{2x_{3}d} using the table of common integrals rule \int a\mathrm{d}x=ax.
\sqrt{2}\sqrt{x_{3}d}x
Simplify.
\begin{matrix}\sqrt{2}\sqrt{x_{3}d}x+С_{3},&\end{matrix}
If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.
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