d) \( \int \frac{\sqrt[5]{1+\sqrt{x}}}{\sqrt{x}} d x \)

To tackle the integral \( \int \frac{\sqrt[5]{1+\sqrt{x}}}{\sqrt{x}} \, dx \), we can utilize a substitution method. Let's set \( u = \sqrt{x} \), implying that \( x = u^2 \) and \( dx = 2u \, du \). Substituting these values into the integral gives: \[ \int \frac{\sqrt[5]{1 + u}}{u} \cdot 2u \, du = 2 \int \sqrt[5]{1 + u} \, du \] Now, the integral becomes simpler: \[ 2 \int \sqrt[5]{1 + u} \, du \] This integral can be solved using a further substitution or applying integration techniques such as integration by parts or recognizing it as a power function. To proceed using a substitution, let \( v = 1 + u \), then \( dv = du \) and \( u = v - 1 \). This changes the integral to: \[ 2 \int (v^{1/5}) \, dv = 2 \cdot \frac{5}{6} v^{6/5} + C = \frac{5}{3} (1 + u)^{6/5} + C \] Re-substituting \( u = \sqrt{x} \) back into the solution, we have: \[ \frac{5}{3} (1 + \sqrt{x})^{6/5} + C \] Thus, the final answer to your integral is: \[ \int \frac{\sqrt[5]{1+\sqrt{x}}}{\sqrt{x}} \, dx = \frac{5}{3} (1 + \sqrt{x})^{6/5} + C \]