\( I = \int \frac { x ^ { 2 } d x } { ( x ^ { 2 } + a ^ { 2 } ) ^ { \frac { 3 } { 2 } } } \)

To solve the integral \[ I = \int \frac{x^2 \, dx}{(x^2 + a^2)^{\frac{3}{2}}} \] we can use a substitution. Let's set \[ u = x^2 + a^2. \] Then, we have \[ du = 2x \, dx \quad \Rightarrow \quad dx = \frac{du}{2x}. \] Also, we can express \(x^2\) in terms of \(u\): \[ x^2 = u - a^2. \] Now, we need to express \(x\) in terms of \(u\): \[ x = \sqrt{u - a^2}. \] Substituting these into the integral, we get: \[ I = \int \frac{(u - a^2) \cdot \frac{du}{2\sqrt{u - a^2}}}{u^{\frac{3}{2}}}. \] This simplifies to: \[ I = \frac{1}{2} \int \frac{(u - a^2)}{u^{\frac{3}{2}} \sqrt{u - a^2}} \, du. \] Now, we can split the integral: \[ I = \frac{1}{2} \left( \int \frac{u}{u^{\frac{3}{2}} \sqrt{u - a^2}} \, du - a^2 \int \frac{1}{u^{\frac{3}{2}} \sqrt{u - a^2}} \, du \right). \] The first integral simplifies to: \[ \int \frac{1}{\sqrt{u - a^2} \sqrt{u}} \, du. \] The second integral can be evaluated using a trigonometric substitution or a standard integral formula. However, a more straightforward approach is to use integration by parts or recognize that the integral can be solved using a known result. The integral \[ \int \frac{x^2}{(x^2 + a^2)^{\frac{3}{2}}} \, dx \] can be evaluated directly using the formula: \[ \int \frac{x^2}{(x^2 + a^2)^{\frac{3}{2}}} \, dx = \frac{x}{\sqrt{x^2 + a^2}} + C. \] Thus, we have: \[ I = \frac{x}{\sqrt{x^2 + a^2}} + C. \] So the final result is: \[ \int \frac{x^2 \, dx}{(x^2 + a^2)^{\frac{3}{2}}} = \frac{x}{\sqrt{x^2 + a^2}} + C. \]