To solve the integral $I = \int x \arctan(4x) \, dx$, we will use integration by parts.

Step 1: Choose $u$ and $dv$

For integration by parts, we use the formula: $\int u \, dv = uv - \int v \, du$

Let:

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

Step 2: Apply the formula

Now substitute into the integration by parts formula: $I = \frac{x^2}{2} \arctan(4x) - \int \frac{x^2}{2} \cdot \frac{4}{1 + 16x^2} \, dx$

Step 3: Simplify the remaining integral

The remaining integral is: $\int \frac{4x^2}{2(1 + 16x^2)} \, dx = 2 \int \frac{x^2}{1 + 16x^2} \, dx$ Now express $x^2$ as $x^2 = \frac{1 + 16x^2 - 1}{16}$, so: $\frac{x^2}{1 + 16x^2} = \frac{1}{16} - \frac{1}{16(1 + 16x^2)}$

Thus, the integral becomes: $2 \int \frac{x^2}{1 + 16x^2} \, dx = 2 \left( \frac{1}{16} \int dx - \frac{1}{16} \int \frac{dx}{1 + 16x^2} \right)$ This simplifies to:

= \frac{x}{8} - \frac{1}{32} \arctan(4x)$$ ### Step 4: Final solution Substitute this back into the original expression for $$I$$: $$I = \frac{x^2}{2} \arctan(4x) - \left( \frac{x}{8} - \frac{1}{32} \arctan(4x) \right)$$ Simplifying: $$I = \frac{x^2}{2} \arctan(4x) - \frac{x}{8} + \frac{1}{32} \arctan(4x) + C$$ Thus, the solution to the integral is: $$\boxed{I = \frac{x^2}{2} \arctan(4x) + \frac{1}{32} \arctan(4x) - \frac{x}{8} + C}$$ Would you like more details on any of the steps? Here are 5 related questions: 1. How is the arctangent function related to inverse trigonometry? 2. Can you explain the steps of integration by parts in more detail? 3. How do you handle integrals involving trigonometric substitution? 4. Why was $$u = \arctan(4x)$$ a good choice for integration by parts? 5. What are the applications of integrals involving inverse trigonometric functions? **Tip:** When doing integration by parts, it helps to choose $$u$$ as a function that simplifies when differentiated and $$dv$$ as something easily integrable.