Is there another way to solve $\int \frac{x}{\sqrt{2x-1}}dx$?

Question

$$\int \frac{x}{\sqrt{2x-1}}dx$$

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?

Answer 1

Another method is to let $u=\sqrt{2x-1}$, so $x=\frac{1}{2}(u^2+1), dx=udu$.

Then $\displaystyle\int\frac{x}{\sqrt{2x-1}}dx=\int\frac{\frac{1}{2}(u^2+1)}{u}\cdot udu=\frac{1}{2}\int(u^2+1)du=\frac{1}{2}\left[\frac{u^3}{3}+u\right]+C$

$\hspace{.23in}=\frac{1}{6}(2x-1)^{3/2}+\frac{1}{2}(2x-1)^{1/2}+C$

Answer 2

Here's a way with trig (though it is harder than what you wrote).

Draw a right triangle with hypotenuse $\sqrt{2x}$. Put a side of length $1$ adjacent to the acute angle $\theta$. Then the opposite side is $\sqrt{2x-1}$. So $x=\sec(\theta)^2/2$,$dx=\sec(\theta)^2\tan(\theta) d \theta$, and $\sqrt{2x-1}=\tan(\theta)$. So you have

$$\frac{1}{2} \int \sec(\theta)^4 d \theta = \frac{1}{2} \int \left ( \sec(\theta)^2 \tan(\theta)^2 + \sec(\theta)^2 \right ) d \theta = \frac{1}{2} \int (u^2+1) du$$

where $u=\tan(\theta)$. Of course, this boils down to $u=\sqrt{2x-1}$, but the steps in the middle were suggested by the trigonometry.

Answer 3

While the approaches already posted work well, here is an approach that provides a straight forward way of deriving the result.

$$\begin{align} \int \frac{x}{\sqrt{2x-1}} dx &= \frac12 \int \frac{2x}{\sqrt{2x-1}} \,\,dx\\\\ &=\frac12 \int \frac{2x-1+1}{\sqrt{2x-1}} \,\,dx\\\\ &=\frac12 \int \sqrt{2x-1}\,\,\,dx+\frac12\int \frac{1}{\sqrt{2x-1}}\,\,\,dx \end{align}$$