We need to evaluate the integral:

$I = \int \frac{-8t^5}{(t^2+2)^{1/2}} \, dt$

Step 1: Substitution

Let: $u = t^2 + 2$ Then, differentiate both sides: $du = 2t \, dt \quad \Rightarrow \quad \frac{du}{2} = t \, dt$

Step 2: Express in Terms of $u$

Rewriting $t^5$ as: $t^5 = t^3 \cdot t^2 = t^3 (u - 2)$

Since $t^3 = t \cdot t^2$, we use $t^2 = u - 2$ and get: $t^3 = t(u - 2)$

Using $t \, dt = \frac{du}{2}$, the integral transforms into:

$I = \int \frac{-8 t^3 (u - 2)}{u^{1/2}} \cdot dt$

Substituting $t^3 = t(u - 2)$:

$I = \int \frac{-8 (u - 2)t}{u^{1/2}} \cdot dt$

Since $t \, dt = \frac{du}{2}$, we substitute:

$I = \int \frac{-8 (u - 2)}{u^{1/2}} \cdot \frac{du}{2}$

$I = \int \frac{-4 (u - 2)}{u^{1/2}} \, du$

Step 3: Split the Integral

$I = -4 \int \frac{u}{u^{1/2}} \, du + 8 \int \frac{1}{u^{1/2}} \, du$

Simplify:

$I = -4 \int u^{1/2} \, du + 8 \int u^{-1/2} \, du$

Step 4: Evaluate the Integrals

Using the power rule $\int u^n \, du = \frac{u^{n+1}}{n+1}$:

$\int u^{1/2} \, du = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2}$

$\int u^{-1/2} \, du = \frac{u^{1/2}}{1/2} = 2u^{1/2}$

So,

$I = -4 \cdot \frac{2}{3} u^{3/2} + 8 \cdot 2 u^{1/2}$

$I = -\frac{8}{3} u^{3/2} + 16 u^{1/2} + C$

Step 5: Substitute Back $u = t^2 + 2$

$I = -\frac{8}{3} (t^2+2)^{3/2} + 16 (t^2+2)^{1/2} + C$

Final Answer:

$\boxed{I = -\frac{8}{3} (t^2+2)^{3/2} + 16 (t^2+2)^{1/2} + C}$

Would you like more details or another approach?

Here are 5 related questions:

1. How do you choose an appropriate substitution for integrals?
2. What is the power rule for integration, and how is it applied?
3. How do you integrate expressions with square roots?
4. How can integration by parts be applied to different functions?
5. How do definite integrals differ from indefinite integrals?

Tip: Always check if substitution simplifies the integral before trying more complex techniques!