To compute the integral of $x \arctan(4x) \, dx$, we will use integration by parts. Recall that integration by parts is given by:

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

Step 1: Choose $u$ and $dv$

Let:

• $u = \arctan(4x)$, so $du = \frac{4}{1 + (4x)^2} dx = \frac{4}{1 + 16x^2} dx$
• $dv = x \, dx$, so $v = \frac{x^2}{2}$

Step 2: Apply the integration by parts formula

$\int x \arctan(4x) \, dx = \frac{x^2}{2} \arctan(4x) - \int \frac{x^2}{2} \cdot \frac{4}{1 + 16x^2} \, dx$

Step 3: Simplify the remaining integral

$\int \frac{4x^2}{2(1 + 16x^2)} \, dx = 2 \int \frac{x^2}{1 + 16x^2} \, dx$

To simplify this, observe that: $\frac{x^2}{1 + 16x^2} = \frac{1}{16} \cdot \frac{16x^2}{1 + 16x^2}$

Now rewrite the remaining integral: $2 \int \frac{x^2}{1 + 16x^2} \, dx = \frac{1}{8} \int \frac{d(1 + 16x^2)}{1 + 16x^2} - \frac{1}{8} \int \frac{1}{1 + 16x^2} \, dx$

This leads to: $\frac{1}{8} \ln(1 + 16x^2) - \frac{1}{32} \arctan(4x)$

Step 4: Combine everything

Thus, the solution is: $\int x \arctan(4x) \, dx = \frac{x^2}{2} \arctan(4x) - \frac{1}{8} \ln(1 + 16x^2) + \frac{1}{32} \arctan(4x) + C$

Final answer:

$\boxed{\frac{x^2}{2} \arctan(4x) - \frac{1}{8} \ln(1 + 16x^2) + \frac{1}{32} \arctan(4x) + C}$

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Would you like more details or have any questions? Here are 5 related questions you might find interesting:

1. How do you apply integration by parts to functions involving logarithms?
2. What is the integral of $\arctan(x)$ without the $x$ factor?
3. Can you use substitution for integrals involving $\arctan$ functions?
4. What is the derivative of $\arctan(x)$?
5. How do you solve integrals with trigonometric substitution?

Tip: Integration by parts is particularly useful when you have a product of functions, such as a polynomial and a logarithmic or inverse trigonometric function.