To solve the definite integral:

$\int_{1}^{4} \frac{1}{\sqrt{x} (\sqrt{x} + 1)^3} dx$

Solution:

1. Substitution: Let $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u \, du$. Also, $\sqrt{x} = u$.

   Update the limits:

   • When $x = 1$, $u = \sqrt{1} = 1$.
   • When $x = 4$, $u = \sqrt{4} = 2$.

   Substituting, the integral becomes: $\int_{1}^{4} \frac{1}{\sqrt{x} (\sqrt{x} + 1)^3} dx = \int_{1}^{2} \frac{1}{u (u + 1)^3} \cdot 2u \, du$

2. Simplify: $\int_{1}^{2} \frac{1}{u (u + 1)^3} \cdot 2u \, du = 2 \int_{1}^{2} \frac{1}{(u + 1)^3} \, du$

3. Integration: For $\frac{1}{(u+1)^3}$, use the formula for $\int u^{-n} du = \frac{u^{-n+1}}{-n+1}$: $\int \frac{1}{(u+1)^3} du = \int (u+1)^{-3} du = \frac{(u+1)^{-2}}{-2} = -\frac{1}{2(u+1)^2}$

4. Apply limits: Now substitute back the limits: $2 \int_{1}^{2} \frac{1}{(u+1)^3} \, du = 2 \left[ -\frac{1}{2(u+1)^2} \right]_1^2$

   Simplify: [ 2 \cdot \left[ -\frac{1}{2(2+1)^2} + \frac{1}{2(1+1)^2} \right] = 2 \cdot \left[ -\frac{1}{18} + \frac{1}{8} \right] ]

5. Common denominator: Combine terms: $-\frac{1}{18} + \frac{1}{8} = \frac{-8 + 18}{144} = \frac{10}{144} = \frac{5}{72}$

6. Final result: Multiply by 2: $2 \cdot \frac{5}{72} = \frac{10}{72} = \frac{5}{36}$

Thus, the value of the definite integral is:

$\boxed{\frac{5}{36}}$

Let me know if you'd like further details or if you have any questions!

Related Questions:

1. How does substitution simplify solving definite integrals?
2. What is the general method for integrating $(x + a)^{-n}$?
3. Can the given integral be solved using numerical methods? Why or why not?
4. What are the applications of definite integrals in physics or engineering?
5. How does changing the integration limits after substitution affect the process?

Tip:

Always recheck limits of integration when using substitution, as it directly impacts the solution!