How do you evaluate the integral #int x/sqrt(4x+1)#?

Note that:

So we can write the integral as:

and integrate by parts:

The resulting integral can be resolved directly using the power rule:

and simplifying:

To evaluate the integral (\int \frac{x}{\sqrt{4x+1}}), you can use a substitution method. Let (u = 4x + 1), then (du = 4dx), or equivalently, (dx = \frac{du}{4}). Substitute these into the integral:

[\int \frac{x}{\sqrt{4x+1}} dx = \int \frac{\frac{u-1}{4}}{\sqrt{u}} \frac{du}{4}]

This simplifies to:

[\frac{1}{16} \int \frac{u-1}{\sqrt{u}} du]

Now, split the integral:

[\frac{1}{16} \left(\int u^{\frac{1}{2}} du - \int u^{-\frac{1}{2}} du\right)]

Integrate each term:

[\frac{1}{16} \left(\frac{2}{3} u^{\frac{3}{2}} - 2 u^{\frac{1}{2}}\right) + C]

Now, resubstitute (u = 4x + 1):

[\frac{1}{16} \left(\frac{2}{3} (4x+1)^{\frac{3}{2}} - 2 (4x+1)^{\frac{1}{2}}\right) + C]

Where (C) is the constant of integration.