What is the value of `int_0^1 (tan^-1 x )/(1+x^2) dx dx`

Options

A. `pi^2/8`
B. `pi^2/32`
C. `pi/4`
D. `pi/8`

Answer

To solve the integral \( I = \int_0^1 \frac{\tan^{-1} x}{1+x^2} \, dx \), we will use substitution.

### Step 1: Substitution
Let \( y = \tan^{-1} x \). Then, we differentiate to find \( dx \):
\[ \frac{dy}{dx} = \frac{1}{1+x^2} \implies dx = (1+x^2) \, dy \]

### Step 2: Change the limits of integration
When \( x = 0 \):
\[ y = \tan^{-1}(0) = 0 \]
When \( x = 1 \):
\[ y = \tan^{-1}(1) = \frac{\pi}{4} \]

### Step 3: Substitute in the integral
Now we substitute \( y \) and \( dx \) into the integral:
\[ I = \int_0^{\frac{\pi}{4}} \frac{y}{1+x^2} (1+x^2) \, dy \]
The \( 1+x^2 \) terms cancel out:
\[ I = \int_0^{\frac{\pi}{4}} y \, dy \]

### Step 4: Evaluate the integral
Now we evaluate the integral:
\[ I = \int_0^{\frac{\pi}{4}} y \, dy = \left[ \frac{y^2}{2} \right]_0^{\frac{\pi}{4}} = \frac{(\frac{\pi}{4})^2}{2} - \frac{0^2}{2} \]
Calculating this gives:
\[ I = \frac{\frac{\pi^2}{16}}{2} = \frac{\pi^2}{32} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{\pi^2}{32} \]