\( 1000\int _{}^{}\frac{x^{3}}{\sqrt {x^{2}+100}} d x \)

To solve the integral \[ 1000 \int \frac{x^3}{\sqrt{x^2 + 100}} \, dx, \] we can use a substitution method. Let's set \[ u = x^2 + 100. \] Then, we differentiate \(u\) with respect to \(x\): \[ du = 2x \, dx \quad \Rightarrow \quad dx = \frac{du}{2x}. \] Next, we express \(x^2\) in terms of \(u\): \[ x^2 = u - 100. \] Now, we can express \(x^3\) in terms of \(u\): \[ x^3 = x \cdot x^2 = x(u - 100). \] To express \(x\) in terms of \(u\), we have: \[ x = \sqrt{u - 100}. \] Thus, \[ x^3 = \sqrt{u - 100} (u - 100). \] Now substituting everything into the integral, we have: \[ 1000 \int \frac{x^3}{\sqrt{x^2 + 100}} \, dx = 1000 \int \frac{\sqrt{u - 100} (u - 100)}{\sqrt{u}} \cdot \frac{du}{2\sqrt{u - 100}}. \] The \(\sqrt{u - 100}\) terms cancel out: \[ = 1000 \int \frac{(u - 100)}{2\sqrt{u}} \, du = 500 \int (u - 100) u^{-1/2} \, du. \] Now we can split the integral: \[ = 500 \left( \int u^{1/2} \, du - 100 \int u^{-1/2} \, du \right). \] Calculating each integral: 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: \[ = 500 \left( \frac{2}{3} u^{3/2} - 100 \cdot 2 u^{1/2} \right) + C, \] which simplifies to: \[ = \frac{1000}{3} u^{3/2} - 1000 u^{1/2} + C. \] Now substituting back \(u = x^2 + 100\): \[ = \frac{1000}{3} (x^2 + 100)^{3/2} - 1000 (x^2 + 100)^{1/2} + C. \] Thus, the final result is: \[ \int \frac{x^3}{\sqrt{x^2 + 100}} \, dx = \frac{1000}{3} (x^2 + 100)^{3/2} - 1000 (x^2 + 100)^{1/2} + C. \]