To determine the indefinite integral of the function [tex]\(\int \left(\frac{6}{t^2} + 8 t^8\right) \, dt\)[/tex], we will integrate each term individually.
### Step-by-Step Integration:
1. Rewrite the integral:
The given integral is:
[tex]\[
\int \left(\frac{6}{t^2} + 8 t^8\right) \, dt
\][/tex]
We can rewrite [tex]\(\frac{6}{t^2}\)[/tex] as [tex]\(6 t^{-2}\)[/tex]. So, the integral becomes:
[tex]\[
\int \left(6 t^{-2} + 8 t^8\right) \, dt
\][/tex]
2. Integrate each term separately:
- For the term [tex]\(6 t^{-2}\)[/tex]:
Applying the power rule for integration, [tex]\(\int t^n \, dt = \frac{t^{n+1}}{n+1} + C\)[/tex], where [tex]\(C\)[/tex] is the constant of integration:
[tex]\[
\int 6 t^{-2} \, dt = 6 \int t^{-2} \, dt = 6 \left( \frac{t^{-2+1}}{-2+1} \right) = 6 \left( \frac{t^{-1}}{-1} \right)
\][/tex]
Simplifying this, we get:
[tex]\[
6 \left( \frac{t^{-1}}{-1} \right) = -6 t^{-1} = -\frac{6}{t}
\][/tex]
- For the term [tex]\(8 t^8\)[/tex]:
Again, using the power rule for integration:
[tex]\[
\int 8 t^8 \, dt = 8 \int t^8 \, dt = 8 \left( \frac{t^{8+1}}{8+1} \right) = 8 \left( \frac{t^9}{9} \right)
\][/tex]
Simplifying this, we get:
[tex]\[
8 \left( \frac{t^9}{9} \right) = \frac{8 t^9}{9}
\][/tex]
3. Combine the results:
Adding the two integrated terms together, we have:
[tex]\[
\int \left(\frac{6}{t^2} + 8 t^8\right) \, dt = -\frac{6}{t} + \frac{8 t^9}{9} + C
\][/tex]
4. Final Answer:
Therefore, the indefinite integral of the given function is:
[tex]\[
\int \left(\frac{6}{t^2} + 8 t^8\right) \, dt = \frac{8 t^9}{9} - \frac{6}{t} + C
\][/tex]
Here, [tex]\(C\)[/tex] represents the constant of integration. This completes our step-by-step solution to the integral.
