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 set: - \( u = \tan^{-1} x \) so that \( du = \frac{1}{1+x^2} \, dx \) - \( dv = x^2 \, dx \) so that \( v = \frac{x^3}{3} \) Now, applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \): \[ \int x^{2} \tan^{-1} x \, dx = \frac{x^3}{3} \tan^{-1} x - \int \frac{x^3}{3} \cdot \frac{1}{1+x^2} \, dx \] Next, we simplify the integral: \[ \int \frac{x^3}{3(1+x^2)} \, dx = \frac{1}{3} \int \frac{x^3}{1+x^2} \, dx \] We can simplify \( \frac{x^3}{1+x^2} = \frac{x^3+x^2-x^2}{1+x^2} = x - \frac{x^2}{1+x^2} \), so we can rewrite the integral: \[ \frac{1}{3} \left( \int x \, dx - \int \frac{x^2}{1+x^2} \, dx \right) \] Evaluating \( \int x \, dx \): \[ \int x \, dx = \frac{x^2}{2} \] Now for \( \int \frac{x^2}{1+x^2} \, dx \): \[ \int \frac{x^2}{1+x^2} \, dx = \int \left(1 - \frac{1}{1+x^2}\right) \, dx = x - \tan^{-1}(x) \] Putting it all together: \[ \int \frac{x^3}{3(1+x^2)} \, dx = \frac{1}{3} \left(\frac{x^2}{2} - \left(x - \tan^{-1}(x)\right)\right) = \frac{x^2}{6} - \frac{1}{3} x + \frac{1}{3} \tan^{-1}(x) \] Finally, substituting back into our equation gives: \[ \int x^{2} \tan^{-1} x \, dx = \frac{x^3}{3} \tan^{-1} x - \left(\frac{x^2}{6} - \frac{1}{3} x + \frac{1}{3} \tan^{-1}(x)\right) + C \] Simplifying further, we have: \[ \int x^{2} \tan^{-1} x \, dx = \frac{x^3}{3} \tan^{-1} x - \frac{x^2}{6} + \frac{1}{3} x - \frac{1}{3} \tan^{-1}(x) + C \] And that gives you the integral \( \int x^{2} \tan^{-1} x \, dx \).