Use any method to evaluate the integral. \[ \int \frac{x^{3} d x}{\sqrt{x^{2}+100}} \]

To evaluate the integral \[ I = \int \frac{x^{3}}{\sqrt{x^{2}+100}} \, dx, \] we can use a combination of substitution and integration by parts. First, let’s perform a substitution. We let \[ u = x^{2} + 100. \] Then, we find its derivative: \[ du = 2x \, dx \quad \Rightarrow \quad dx = \frac{du}{2x}. \] Next, we need to express \(x^3\) in terms of \(u\). Since \(u = x^{2} + 100\), we have \[ x^{2} = u - 100 \quad \Rightarrow \quad x^{3} = x \cdot x^{2} = x(u - 100). \] Now we substitute \(x\) back in terms of \(u\). We also know that \(x = \sqrt{u - 100}\). Thus, \[ x^{3} = \sqrt{u - 100} (u - 100). \] With these substitutions, the integral becomes \[ I = \int \frac{\sqrt{u - 100} (u - 100)}{\sqrt{u}} \cdot \frac{du}{2\sqrt{u - 100}}. \] This simplifies to \[ I = \frac{1}{2} \int (u - 100) \sqrt{u} \, du. \] Now, we can expand \((u - 100)\sqrt{u}\): \[ (u - 100)\sqrt{u} = u^{3/2} - 100\sqrt{u}. \] Thus, the integral becomes \[ I = \frac{1}{2} \left( \int u^{3/2} \, du - 100 \int u^{1/2} \, du \right). \] We can now evaluate these integrals separately: 1. The integral of \(u^{3/2}\): \[ \int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}. \] 2. The integral of \(u^{1/2}\): \[ \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}. \] Now substituting back: \[ I = \frac{1}{2} \left( \frac{2}{5} u^{5/2} - 100 \cdot \frac{2}{3} u^{3/2} \right). \] Simplifying this gives: \[ I = \frac{1}{5} u^{5/2} - \frac{100}{3} u^{3/2}. \] Substituting back \(u = x^{2} + 100\): \[ I = \frac{1}{5} (x^{2}+100)^{5/2} - \frac{100}{3} (x^{2}+100)^{3/2} + C. \] Thus, the final result for the integral is \[ \int \frac{x^{3}}{\sqrt{x^{2}+100}} \, dx = \frac{1}{5} (x^{2}+100)^{5/2} - \frac{100}{3} (x^{2}+100)^{3/2} + C. \]