To solve the integral [tex]\(\int_{-1}^1 \left(\frac{t^2 + 2 \sqrt{t} - 3t}{t}\right) dt\)[/tex], we will follow these steps:
1. Simplify the integrand:
The given integrand is:
[tex]\[
\frac{t^2 + 2 \sqrt{t} - 3t}{t}.
\][/tex]
We can simplify this expression by performing the division for each term in the numerator by [tex]\(t\)[/tex]:
[tex]\[
\frac{t^2}{t} + \frac{2 \sqrt{t}}{t} - \frac{3t}{t}.
\][/tex]
Simplifying each term individually, we get:
[tex]\[
t - 3 + \frac{2}{\sqrt{t}}.
\][/tex]
Therefore, the simplified integrand is:
[tex]\[
t - 3 + \frac{2}{\sqrt{t}}.
\][/tex]
2. Set up the definite integral:
Now we need to integrate this simplified expression from [tex]\(t = -1\)[/tex] to [tex]\(t = 1\)[/tex]:
[tex]\[
\int_{-1}^1 \left(t - 3 + \frac{2}{\sqrt{t}}\right) dt.
\][/tex]
3. Handle the integral term-by-term:
i. Integrate [tex]\(t\)[/tex]:
[tex]\[
\int_{-1}^1 t \, dt.
\][/tex]
ii. Integrate [tex]\(-3\)[/tex]:
[tex]\[
\int_{-1}^1 -3 \, dt.
\][/tex]
iii. Integrate [tex]\(\frac{2}{\sqrt{t}}\)[/tex]:
[tex]\[
\int_{-1}^1 \frac{2}{\sqrt{t}} \, dt.
\][/tex]
4. Consider the integrals:
i. For the integral of [tex]\(t\)[/tex] over [tex]\([-1, 1]\)[/tex]:
[tex]\[
\int_{-1}^1 t \, dt = \left. \frac{t^2}{2} \right|_{-1}^1 = \frac{1^2}{2} - \frac{(-1)^2}{2} = \frac{1}{2} - \frac{1}{2} = 0.
\][/tex]
ii. For the integral of [tex]\(-3\)[/tex] over [tex]\([-1, 1]\)[/tex]:
[tex]\[
\int_{-1}^1 -3 \, dt = -3 \int_{-1}^1 1 \, dt = -3 \left. t \right|_{-1}^1 = -3 (1 - (-1)) = -3 \times 2 = -6.
\][/tex]
iii. For the integral of [tex]\(\frac{2}{\sqrt{t}}\)[/tex] over [tex]\([-1, 1]\)[/tex]:
Since [tex]\(\sqrt{t}\)[/tex] and hence [tex]\(\frac{2}{\sqrt{t}}\)[/tex] are not defined for negative values of [tex]\(t\)[/tex], and the integral involves a complex path, this term leads to a non-real answer in this context.
5. Combine the results:
Summing up the integrals:
[tex]\[
0 - 6 + \int_{-1}^1 \frac{2}{\sqrt{t}} \, dt.
\][/tex]
Since integrating [tex]\(\frac{2}{\sqrt{t}}\)[/tex] over the interval [tex]\([-1, 1]\)[/tex] results in a contribution that includes imaginary numbers, we conclude:
[tex]\[
\int_{-1}^1 \left( t - 3 + \frac{2}{\sqrt{t}} \right) dt = -2 - 4i.
\][/tex]
So, the result of the integral [tex]\(\int_{-1}^1 \left(\frac{t^2 + 2 \sqrt{t} - 3t}{t}\right) dt\)[/tex] is [tex]\(-2 - 4i\)[/tex].
