The given integral is:

$\int_{\sqrt{2}}^{1} \left( \frac{u^5}{4} - \frac{1}{u^3} \right) \, du$

Step 1: Break the integral into parts

$\int_{\sqrt{2}}^{1} \left( \frac{u^5}{4} - \frac{1}{u^3} \right) \, du = \frac{1}{4} \int_{\sqrt{2}}^{1} u^5 \, du - \int_{\sqrt{2}}^{1} \frac{1}{u^3} \, du$

Step 2: Solve each part of the integral

(a) First term: $\frac{1}{4} \int u^5 \, du$

Using the power rule for integration: $\int u^n \, du = \frac{u^{n+1}}{n+1}, \quad n \neq -1$

Here, $n = 5$: $\int u^5 \, du = \frac{u^{6}}{6}$ So: $\frac{1}{4} \int u^5 \, du = \frac{1}{4} \cdot \frac{u^{6}}{6} = \frac{u^{6}}{24}$

(b) Second term: $\int \frac{1}{u^3} \, du$

Rewrite $\frac{1}{u^3}$ as $u^{-3}$: $\int u^{-3} \, du = \frac{u^{-3+1}}{-3+1} = \frac{u^{-2}}{-2} = -\frac{1}{2u^2}$

Step 3: Combine results

The integral becomes: [ \int_{\sqrt{2}}^{1} \left( \frac{u^5}{4} - \frac{1}{u^3} \right) , du = \left[ \frac{u^6}{24} \right]{\sqrt{2}}^{1} - \left[ -\frac{1}{2u^2} \right]{\sqrt{2}}^{1} ]

Step 4: Evaluate definite integrals

(a) For $\frac{u^6}{24}$:

$\left[ \frac{u^6}{24} \right]_{\sqrt{2}}^{1} = \frac{1^6}{24} - \frac{(\sqrt{2})^6}{24}$ $= \frac{1}{24} - \frac{(2^3)}{24} = \frac{1}{24} - \frac{8}{24} = -\frac{7}{24}$

(b) For $-\frac{1}{2u^2}$:

$\left[ -\frac{1}{2u^2} \right]_{\sqrt{2}}^{1} = -\frac{1}{2(1^2)} + \frac{1}{2(\sqrt{2})^2}$ $= -\frac{1}{2} + \frac{1}{2(2)} = -\frac{1}{2} + \frac{1}{4} = -\frac{2}{4} + \frac{1}{4} = -\frac{1}{4}$

Step 5: Combine results

The total integral is: $-\frac{7}{24} - \frac{1}{4}$ Convert $-\frac{1}{4}$ to have a denominator of 24: $-\frac{7}{24} - \frac{6}{24} = -\frac{13}{24}$

Final Answer:

$\boxed{-\frac{13}{24}}$

Let me know if you would like more details or have any questions!

Related Questions:

1. How do you apply the power rule to integrals?
2. What are the common rules for definite integrals with fractional powers?
3. How can substitution simplify similar integrals?
4. What happens when $n = -1$ in $\int u^n \, du$?
5. How can this integral be solved using numerical methods instead of symbolic integration?

Tip:

Always double-check the limits of definite integrals, as switching the limits can change the sign of the result!