Questions: ∫ x / √(2x - 1) dx

Solution Steps

To solve the integral \(\int \frac{x}{\sqrt{2x-1}} \, dx\), we can use a substitution method. Let \(u = 2x - 1\), then \(du = 2dx\). This substitution will simplify the integral into a more manageable form.

We start with the integral

\[ I = \int \frac{x}{\sqrt{2x - 1}} \, dx. \]

Let

\[ u = 2x - 1 \implies du = 2dx \implies dx = \frac{du}{2}. \]

Rewriting \(x\) in terms of \(u\):

\[ x = \frac{u + 1}{2}. \]

Substituting these into the integral gives:

\[ I = \int \frac{\frac{u + 1}{2}}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{4} \int \frac{u + 1}{\sqrt{u}} \, du = \frac{1}{4} \int \left( u^{1/2} + u^{-1/2} \right) \, du. \]

Now we can integrate:

\[ \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}, \] \[ \int u^{-1/2} \, du = 2u^{1/2}. \]

Thus, we have:

\[ I = \frac{1}{4} \left( \frac{2}{3} u^{3/2} + 2u^{1/2} \right) + C = \frac{1}{6} u^{3/2} + \frac{1}{2} u^{1/2} + C. \]

Substituting back \(u = 2x - 1\):

\[ I = \frac{1}{6} (2x - 1)^{3/2} + \frac{1}{2} (2x - 1)^{1/2} + C. \]

The integral can be expressed as a piecewise function:

\[ I = \begin{cases} \frac{x \sqrt{2x - 1}}{3} + \frac{\sqrt{2x - 1}}{3} & \text{if } |x| > \frac{1}{2} \\ \frac{I x \sqrt{1 - 2x}}{3} + \frac{I \sqrt{1 - 2x}}{3} & \text{otherwise} \end{cases}. \]

Final Answer