Evaluate `int_(r_(1))^(r_(2))(r- eta (q_(0)q)/(r^(2)))dr`

Answer

To evaluate the integral

\[ I = \int_{r_1}^{r_2} \left( r - \eta \frac{q_0 q}{r^2} \right) dr \]

we can break it down into two separate integrals:

\[ I = \int_{r_1}^{r_2} r \, dr - \eta q_0 q \int_{r_1}^{r_2} \frac{1}{r^2} \, dr \]

### Step 1: Evaluate the first integral \(\int_{r_1}^{r_2} r \, dr\)

The integral of \(r\) is:

\[ \int r \, dr = \frac{r^2}{2} \]

Now we will evaluate this from \(r_1\) to \(r_2\):

\[ \int_{r_1}^{r_2} r \, dr = \left[ \frac{r^2}{2} \right]_{r_1}^{r_2} = \frac{r_2^2}{2} - \frac{r_1^2}{2} = \frac{r_2^2 - r_1^2}{2} \]

### Step 2: Evaluate the second integral \(\int_{r_1}^{r_2} \frac{1}{r^2} \, dr\)

The integral of \(\frac{1}{r^2}\) is:

\[ \int \frac{1}{r^2} \, dr = -\frac{1}{r} \]

Now we will evaluate this from \(r_1\) to \(r_2\):

\[ \int_{r_1}^{r_2} \frac{1}{r^2} \, dr = \left[-\frac{1}{r}\right]_{r_1}^{r_2} = -\frac{1}{r_2} + \frac{1}{r_1} = \frac{1}{r_1} - \frac{1}{r_2} \]

### Step 3: Combine the results

Now we can substitute the results of the two integrals back into our expression for \(I\):

\[ I = \frac{r_2^2 - r_1^2}{2} - \eta q_0 q \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]

### Final Result

Thus, the final result for the integral is:

\[ I = \frac{r_2^2 - r_1^2}{2} - \eta q_0 q \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]