arctan(√6x−1)




Calculate indefinite integrals

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.

Learn how to calculate the indefinite integral of arctan(sqrt(6x - 1)) using integration by parts with a step-by-step solution. This problem involves integrating the arctangent function and applying the integration by parts formula. Suitable for college-level students and those interested in calculus.

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