Question

Evaluate: \[ \int x \tan^{-1}{x} \, dx \]

Hint:

Use integration by parts to handle integrals involving products of functions like \( x \) and \( \tan^{-1}{x} \).

Answer 1

We use integration by parts. Let: \[ u = \tan^{-1}{x}, \quad dv = x \, dx \] Then: \[ du = \frac{1}{1+x^2} \, dx, \quad v = \frac{x^2}{2} \] Step 1: Apply the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Step 2: Substitute: \[ \int x \tan^{-1}{x} \, dx = \frac{x^2}{2} \tan^{-1}{x} - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx \] Step 3: Simplify the second integral: \[ \int \frac{x^2}{2(1+x^2)} \, dx = \int \frac{1}{2} \, dx = \frac{x}{2} \] Step 4: Final result: \[ \int x \tan^{-1}{x} \, dx = \frac{x^2}{2} \tan^{-1}{x} - \frac{x}{2} + C \] Thus, the integral is: \[ \boxed{\frac{x^2}{2} \tan^{-1}{x} - \frac{x}{2} + C} \]