`int_(0)^(1) ( tan^(-1)x)/( 1+x^(2)) dx` is equal to

Options

A. `(pi)/( 4)`
B. `(pi^(2))/( 32)`
C. 1
D. `(pi^(2))/( 16)`

Answer

To solve the integral \( I = \int_{0}^{1} \frac{\tan^{-1} x}{1+x^2} \, dx \), we can follow these steps:

### Step 1: Define the Integral
Let
\[ I = \int_{0}^{1} \frac{\tan^{-1} x}{1+x^2} \, dx \]

### Step 2: Substitution
We notice that the derivative of \( \tan^{-1} x \) is \( \frac{1}{1+x^2} \). Therefore, we can use the substitution:
\[ t = \tan^{-1} x \implies dt = \frac{1}{1+x^2} \, dx \]
This implies that:
\[ dx = (1+x^2) \, dt \]

### Step 3: Change the Limits
Next, we need to change the limits of integration. When \( x = 0 \):
\[ t = \tan^{-1}(0) = 0 \]
When \( x = 1 \):
\[ t = \tan^{-1}(1) = \frac{\pi}{4} \]
Thus, the limits change from \( x: 0 \to 1 \) to \( t: 0 \to \frac{\pi}{4} \).

### Step 4: Rewrite the Integral
Substituting these into the integral gives:
\[ I = \int_{0}^{\frac{\pi}{4}} t \, dt \]

### Step 5: Integrate
Now we can integrate:
\[ I = \left[ \frac{t^2}{2} \right]_{0}^{\frac{\pi}{4}} = \frac{(\frac{\pi}{4})^2}{2} - \frac{0^2}{2} \]
Calculating this gives:
\[ I = \frac{\frac{\pi^2}{16}}{2} = \frac{\pi^2}{32} \]

### Step 6: Conclusion
Thus, the value of the integral is:
\[ \boxed{\frac{\pi^2}{32}} \]

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