\( \int _{}^{}\frac{x^{3}}{\sqrt {x^{2}+100}} d x \)

To solve the integral \[ I = \int \frac{x^3}{\sqrt{x^2 + 100}} \, dx, \] we can use the substitution \( u = x^2 + 100 \). Then, we have: \[ du = 2x \, dx \quad \Rightarrow \quad dx = \frac{du}{2x}. \] Also, from the substitution, we can express \( x^2 \) in terms of \( u \): \[ x^2 = u - 100 \quad \Rightarrow \quad x^3 = x \cdot x^2 = x(u - 100). \] Now, we need to express \( x \) in terms of \( u \). From \( u = x^2 + 100 \), we have: \[ x = \sqrt{u - 100}. \] Substituting \( x \) and \( dx \) into the integral, we get: \[ I = \int \frac{(\sqrt{u - 100})(u - 100)}{\sqrt{u}} \cdot \frac{du}{2\sqrt{u - 100}}. \] The \( \sqrt{u - 100} \) terms cancel out: \[ I = \int \frac{(u - 100)}{2\sqrt{u}} \, du. \] Now, we can split the integral: \[ I = \frac{1}{2} \int \frac{u}{\sqrt{u}} \, du - \frac{100}{2} \int \frac{1}{\sqrt{u}} \, du. \] This simplifies to: \[ I = \frac{1}{2} \int u^{1/2} \, du - 50 \int u^{-1/2} \, du. \] Now we can integrate each term: 1. For \( \int u^{1/2} \, du \): \[ \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}. \] 2. For \( \int u^{-1/2} \, du \): \[ \int u^{-1/2} \, du = 2u^{1/2}. \] Putting it all together, we have: \[ I = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} - 50 \cdot 2 u^{1/2} + C, \] which simplifies to: \[ I = \frac{1}{3} u^{3/2} - 100 u^{1/2} + C. \] Now, substituting back \( u = x^2 + 100 \): \[ I = \frac{1}{3} (x^2 + 100)^{3/2} - 100 (x^2 + 100)^{1/2} + C. \] Thus, the final result for the integral is: \[ \int \frac{x^3}{\sqrt{x^2 + 100}} \, dx = \frac{1}{3} (x^2 + 100)^{3/2} - 100 (x^2 + 100)^{1/2} + C. \]