`int(t-cos omegat+(1)/(t))dt`

Options

A. `(t^(2))/(2)-(sin omegat)/(omega)+t`
B. `(t^(2))/(2)-(sin omegat)/(omega)+logt`.
C. `(t^(2))/(2)-(sin omegat)+logt`.
D. `(t^(2))/(2)-(cos omegat)/(omega)+logt.`

Answer

To solve the integral \( \int \left( t - \cos(\omega t) + \frac{1}{t} \right) dt \), we can break it down into three separate integrals:

1. **Integrate \( t \)**:
\[ \int t \, dt = \frac{t^2}{2} \]

2. **Integrate \( -\cos(\omega t) \)**:
To integrate \( -\cos(\omega t) \), we use the formula for the integral of cosine:
\[ \int -\cos(\omega t) \, dt = -\frac{1}{\omega} \sin(\omega t) \]

3. **Integrate \( \frac{1}{t} \)**:
The integral of \( \frac{1}{t} \) is:
\[ \int \frac{1}{t} \, dt = \ln |t| \]

Now, we can combine all these results together:

\[ \int \left( t - \cos(\omega t) + \frac{1}{t} \right) dt = \frac{t^2}{2} - \frac{1}{\omega} \sin(\omega t) + \ln |t| + C \]

where \( C \) is the constant of integration.

Thus, the final answer is:
\[ \frac{t^2}{2} - \frac{1}{\omega} \sin(\omega t) + \ln |t| + C \]