\( \int \frac { \tan ^ { - 1 } x } { 1 + \frac { 1 } { x ^ { 2 } } } d x \)

To solve the integral \( \int \frac { \tan ^ { - 1 } x } { 1 + \frac { 1 } { x ^ { 2 } } } d x \), let's first simplify the integrand. The expression simplifies as follows: \[ 1 + \frac {1}{x^2} = \frac{x^2 + 1}{x^2} \] Thus, the integral can be rewritten as: \[ \int \tan^{-1}(x) \cdot \frac{x^2}{x^2 + 1} \, dx \] Now, we can perform integration by parts, letting \( u = \tan^{-1}(x) \) and \( dv = \frac{x^2}{x^2 + 1} \, dx \). Then we find \( du \) and \( v \): \[ du = \frac{1}{1 + x^2} \, dx \] For \( v \), we can integrate \( dv \): \[ dv = \frac{x^2}{x^2 + 1} \, dx = \left( 1 - \frac{1}{x^2 + 1} \right) \, dx \] Thus, \[ v = x - \tan^{-1}(x) \] Now we can apply integration by parts: \[ \int u \, dv = uv - \int v \, du \] This leads to: \[ \int \tan^{-1}(x) \cdot \frac{x^2}{x^2 + 1} \, dx = \tan^{-1}(x)(x - \tan^{-1}(x)) - \int (x - \tan^{-1}(x)) \cdot \frac{1}{1+x^2} \, dx \] This integral can be evaluated, resulting in: \[ \frac{1}{2} \tan^{-1}^2(x) + C + \ln(x^2 + 1) \] Thus, the final result, after solving, is: \[ \frac{x}{2}( \tan^{-1}(x) - \frac{1}{2}\ln(x^2 + 1) ) + C \]