The value of `int_1^e((tan^(-1)x)/x+(logx)/(1+x^2))dxi s` `tane` (b) `tan^(-1)e` `tan^(-1)(1/e)` (d) none of these

Options

A. `tane`
B. `tan^(-1)e`
C. `tan^(-1)(1//e)`
D. none of these

Explanation

To solve the integral

\[ I = \int_1^e \left( \frac{\tan^{-1} x}{x} + \frac{\log x}{1+x^2} \right) dx, \]

we can break it down into two separate integrals:

\[ I = \int_1^e \frac{\tan^{-1} x}{x} \, dx + \int_1^e \frac{\log x}{1+x^2} \, dx. \]

Let's denote these two integrals as \( I_1 \) and \( I_2 \):

\[ I_1 = \int_1^e \frac{\tan^{-1} x}{x} \, dx, \]
\[ I_2 = \int_1^e \frac{\log x}{1+x^2} \, dx. \]

### Step 1: Evaluate \( I_2 \) using Integration by Parts

For \( I_2 \), we will use integration by parts. Let:

- \( u = \log x \) (then \( du = \frac{1}{x} \, dx \))
- \( dv = \frac{1}{1+x^2} \, dx \) (then \( v = \tan^{-1} x \))

Using the integration by parts formula:

\[ \int u \, dv = uv - \int v \, du, \]

we have:

\[ I_2 = \left[ \log x \cdot \tan^{-1} x \right]_1^e - \int_1^e \tan^{-1} x \cdot \frac{1}{x} \, dx. \]

### Step 2: Evaluate the Boundary Terms

Now we evaluate the boundary terms:

\[ \left[ \log x \cdot \tan^{-1} x \right]_1^e = \log e \cdot \tan^{-1} e - \log 1 \cdot \tan^{-1} 1 = 1 \cdot \tan^{-1} e - 0 \cdot \frac{\pi}{4} = \tan^{-1} e. \]

### Step 3: Combine the Results

Now substituting back into the expression for \( I_2 \):

\[ I_2 = \tan^{-1} e - I_1. \]

### Step 4: Substitute \( I_2 \) Back into \( I \)

Now substituting \( I_2 \) back into the expression for \( I \):

\[ I = I_1 + I_2 = I_1 + \tan^{-1} e - I_1 = \tan^{-1} e. \]

### Final Result

Thus, the value of the integral is:

\[ I = \tan^{-1} e. \]