\( \int _{}^{}\frac{\sqrt {x}}{7x^{5}} d x \)
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To solve the integral \( \int \frac{\sqrt{x}}{7x^5} \, dx \), we first simplify the integrand: \[ \frac{\sqrt{x}}{7x^5} = \frac{x^{1/2}}{7x^5} = \frac{1}{7} x^{1/2 - 5} = \frac{1}{7} x^{-9/2} \] Now we can rewrite the integral: \[ \int \frac{\sqrt{x}}{7x^5} \, dx = \frac{1}{7} \int x^{-9/2} \, dx \] Using the power rule for integration, we get: \[ \int x^{-9/2} \, dx = \frac{x^{-7/2}}{-7/2} + C = -\frac{2}{7} x^{-7/2} + C \] Substituting this back into the integral gives: \[ \int \frac{\sqrt{x}}{7x^5} \, dx = -\frac{2}{7} \cdot \frac{1}{7} x^{-7/2} + C = -\frac{2}{49} x^{-7/2} + C \] Thus, the final answer is: \[ -\frac{2}{49} x^{-7/2} + C \]