\( \int \frac { 11 } { \sqrt { x ^ { 3 } + 100 } } d x \)

To solve the integral \[ \int \frac{11}{\sqrt{x^3 + 100}} \, dx, \] we can use a substitution to simplify the expression under the square root. Let's set \[ u = x^3 + 100. \] Then, we differentiate \(u\) with respect to \(x\): \[ \frac{du}{dx} = 3x^2 \implies du = 3x^2 \, dx \implies dx = \frac{du}{3x^2}. \] Next, we need to express \(x^2\) in terms of \(u\). From our substitution, we have: \[ x^3 = u - 100 \implies x = (u - 100)^{1/3}. \] Thus, \[ x^2 = \left((u - 100)^{1/3}\right)^2 = (u - 100)^{2/3}. \] Now we can substitute \(dx\) and \(x^2\) into the integral: \[ \int \frac{11}{\sqrt{u}} \cdot \frac{du}{3(u - 100)^{2/3}}. \] This gives us: \[ \int \frac{11}{3} \cdot \frac{1}{(u - 100)^{2/3} \sqrt{u}} \, du. \] This integral is quite complex, so let's revert to the original integral and try a different approach. We can factor out constants and use a trigonometric substitution. We can rewrite the integral as: \[ 11 \int \frac{1}{\sqrt{x^3 + 100}} \, dx. \] Now, we can use a trigonometric substitution. Let: \[ x^3 + 100 = 100 \sec^2(\theta). \] Then, \[ x^3 = 100 \sec^2(\theta) - 100 = 100(\sec^2(\theta) - 1) = 100 \tan^2(\theta). \] Taking the cube root gives: \[ x = 10 \tan(\theta). \] Now, we differentiate \(x\): \[ dx = 10 \sec^2(\theta) \, d\theta. \] Substituting into the integral, we have: \[ \int \frac{11}{\sqrt{100 \sec^2(\theta)}} \cdot 10 \sec^2(\theta) \, d\theta = \int \frac{11 \cdot 10 \sec^2(\theta)}{10 \sec(\theta)} \, d\theta = \int 11 \sec(\theta) \, d\theta. \] The integral of \(\sec(\theta)\) is: \[ \int \sec(\theta) \, d\theta = \ln | \sec(\theta) + \tan(\theta) | + C. \] Thus, we have: \[ 11 \int \sec(\theta) \, d\theta = 11 \ln | \sec(\theta) + \tan(\theta) | + C. \] Now we need to convert back to \(x\). Recall that: \[ \tan(\theta) = \frac{x}{10} \quad \text{and} \quad \sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \sqrt{1 + \left(\frac{x}{10}\right)^2} = \frac{\sqrt{x^2 + 100}}{10}. \] Thus, \[ \sec(\theta) + \tan(\theta) = \frac{\sqrt{x^2 + 100}}{10} + \frac{x}{10} = \frac{\sqrt{x^2 + 100} + x}{10}. \] Putting it all together, we have: \[ \int \frac{11}{\sqrt{x^3 + 100}} \, dx = 11 \ln \left| \frac{\sqrt{x^2 + 100} + x}{10} \right| + C. \] Finally, we can simplify this to: \[ \int \frac{11}{\sqrt{x^3 + 100}} \, dx = 11 \ln \left( \sqrt{x^2 + 100} + x \right) - 11 \ln(10) + C. \] Thus, the final answer is: \[ \int \frac{11}{\sqrt{x^3 + 100}} \, dx = 11 \ln \left( \sqrt{x^2