To find the indefinite integral of $\arctan(\sqrt{6x - 1})$, we will use integration by parts. The integration by parts formula is:

$\int u \, dv = uv - \int v \, du$

Let's set $u = \arctan(\sqrt{6x - 1})$ and $dv = dx$.

First, we need to find $du$ and $v$:

1. Finding $du$:

   $u = \arctan(\sqrt{6x - 1})$

   Let $t = \sqrt{6x - 1}$, then:

   $u = \arctan(t)$

   $\frac{dt}{dx} = \frac{6}{2\sqrt{6x - 1}} = \frac{3}{\sqrt{6x - 1}}$

   Therefore,

   $\frac{du}{dx} = \frac{d}{dx} \arctan(t) \cdot \frac{dt}{dx} = \frac{1}{1 + t^2} \cdot \frac{3}{\sqrt{6x - 1}}$

   Substituting back $t = \sqrt{6x - 1}$:

   $\frac{du}{dx} = \frac{1}{1 + (\sqrt{6x - 1})^2} \cdot \frac{3}{\sqrt{6x - 1}} = \frac{1}{1 + 6x - 1} \cdot \frac{3}{\sqrt{6x - 1}} = \frac{3}{6x} = \frac{1}{2x}$

   Hence,

   $du = \frac{1}{2x} dx$

2. Finding $v$:

   $dv = dx \implies v = x$

Now, applying the integration by parts formula:

$\int \arctan(\sqrt{6x - 1}) \, dx = x \arctan(\sqrt{6x - 1}) - \int x \cdot \frac{1}{2x} \, dx$

Simplify the integral on the right-hand side:

$\int x \cdot \frac{1}{2x} \, dx = \int \frac{1}{2} \, dx = \frac{1}{2} \int dx = \frac{1}{2} x$

Putting it all together:

$\int \arctan(\sqrt{6x - 1}) \, dx = x \arctan(\sqrt{6x - 1}) - \frac{1}{2} x + C$

Therefore, the indefinite integral of $\arctan(\sqrt{6x - 1})$ is:

$\boxed{x \arctan(\sqrt{6x - 1}) - \frac{1}{2} x + C}$

Would you like further details or have any questions?

Here are 5 more questions you might be interested in:

1. How do you find the indefinite integral of $e^{2x}$?
2. What is the derivative of $\ln(x)$?
3. How do you solve the integral $\int \sin^2(x) \, dx$?
4. What is the method of partial fractions in integration?
5. How do you evaluate the definite integral of $\int_0^\pi \sin(x) \, dx$?

Tip: Integration by parts is particularly useful for integrals involving products of functions, where you can identify one part to differentiate and another to integrate.