Use integral tables to evaluate the integral. \[ \int x^{2} \tan ^{-1} x d x \]

To evaluate the integral \( \int x^{2} \tan^{-1} x \, dx \), we can use integration by parts. We will let: - \( u = \tan^{-1} x \) so that \( du = \frac{1}{1+x^2} \, dx \) - \( dv = x^2 \, dx \) so that \( v = \frac{x^3}{3} \) Applying integration by parts, we use the formula: \[ \int u \, dv = uv - \int v \, du \] We substitute in our choices: \[ \int x^{2} \tan^{-1} x \, dx = \tan^{-1} x \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{1+x^2} \, dx \] This simplifies to: \[ \int x^{2} \tan^{-1} x \, dx = \frac{x^3}{3} \tan^{-1} x - \frac{1}{3} \int \frac{x^3}{1+x^2} \, dx \] Next, we simplify the integral \(\int \frac{x^3}{1+x^2} \, dx\): \[ \frac{x^3}{1+x^2} = \frac{x^3 + x^2 - x^2}{1+x^2} = x - \frac{x^2}{1+x^2} \] Then, we can write: \[ \int \frac{x^3}{1+x^2} \, dx = \int x \, dx - \int \frac{x^2}{1+x^2} \, dx \] Calculating the first integral: \[ \int x \, dx = \frac{x^2}{2} \] Now calculate \(\int \frac{x^2}{1+x^2} \, dx\), which can be approached using substitution or rewriting: \[ \int \frac{x^2}{1+x^2} \, dx = \int (1 - \frac{1}{1+x^2}) \, dx = x - \tan^{-1} x \] Putting it all together: \[ \int \frac{x^3}{1+x^2} \, dx = \frac{x^2}{2} - \left( x - \tan^{-1} x \right) \] \[ = \frac{x^2}{2} - x + \tan^{-1} x \] Now substituting back into our expression for \(\int x^{2} \tan^{-1} x \, dx\): \[ \int x^{2} \tan^{-1} x \, dx = \frac{x^3}{3} \tan^{-1} x - \frac{1}{3} \left( \frac{x^2}{2} - x + \tan^{-1} x \right) \] Distributing \(-\frac{1}{3}\): \[ = \frac{x^3}{3} \tan^{-1} x - \frac{x^2}{6} + \frac{x}{3} - \frac{1}{3} \tan^{-1} x \] Finally, combining like terms gives: \[ = \left( \frac{x^3}{3} - \frac{1}{3} \right) \tan^{-1} x + \frac{x}{3} - \frac{x^2}{6} + C \] Thus, the evaluated integral is: \[ \int x^{2} \tan^{-1} x \, dx = \frac{x^3}{3} \tan^{-1} x - \frac{x^2}{6} + \frac{x}{3} + C \]