<s>[INST] Is there another way to solve $\int \frac{x}{\sqrt{2x-1}}dx$? $$\int \frac{x}{\sqrt{2x-1}}dx$$
Let $u=2x-1$
$du=2dx$
$$=\frac{1}{2}\int \frac{u+1}{2\sqrt{u}}du$$
$$=\frac{1}{2}\int (\frac{\sqrt{u}}{2}+\frac{1}{2\sqrt{u}})du$$
$$=\frac{1}{4}\int \sqrt{u}du+\frac{1}{4}\int\frac{1}{\sqrt{u}}du$$
$$=\frac{u^{\frac{3}{2}}}{6}+\frac{\sqrt{u}}{2}+c$$
$$=\frac{1}{6}(2x-1)^{\frac{3}{2}}+\frac{1}{2}\sqrt{2x-1}+c$$
$$=\frac{1}{3}(x+1)\sqrt{2x-1}+c$$
Is there another way except this?
 [/INST] While the approaches already posted work well, here is an approach that provides a straight forward way of deriving the result.
$$\begin{align}
\int \frac{x}{\sqrt{2x-1}} dx &= \frac12 \int \frac{2x}{\sqrt{2x-1}} \,\,dx\\\\
&=\frac12 \int \frac{2x-1+1}{\sqrt{2x-1}} \,\,dx\\\\
&=\frac12 \int \sqrt{2x-1}\,\,\,dx+\frac12\int \frac{1}{\sqrt{2x-1}}\,\,\,dx
\end{align}$$
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