Area bounded by the curve `y = tan^(-1)x`, the X-axis and the line x = 1 is

Options

A. `(pi)/(4)+log sqrt(2)`
B. `(pi)/(4)-sqrt(2)`
C. `(pi)/(4)-log sqrt(2)`
D. `(pi)/(2)-log sqrt(2)`

Answer

To find the area bounded by the curve \( y = \tan^{-1}x \), the X-axis, and the line \( x = 1 \), we can follow these steps:

### Step 1: Identify the area to be calculated
The area we need to calculate is bounded by the curve \( y = \tan^{-1}x \), the X-axis (which is \( y = 0 \)), and the vertical line \( x = 1 \).

### Step 2: Set up the integral
The area \( A \) can be expressed as the integral of the function \( y = \tan^{-1}x \) from \( x = 0 \) to \( x = 1 \):
\[ A = \int_{0}^{1} \tan^{-1}x \, dx \]

### Step 3: Integrate the function
To integrate \( \tan^{-1}x \), we can use integration by parts. Let:
- \( u = \tan^{-1}x \) ⇒ \( du = \frac{1}{1+x^2} \, dx \)
- \( dv = dx \) ⇒ \( v = x \)

Using integration by parts, we have:
\[ \int u \, dv = uv - \int v \, du \]
Thus,
\[ \int \tan^{-1}x \, dx = x \tan^{-1}x - \int x \cdot \frac{1}{1+x^2} \, dx \]

### Step 4: Solve the remaining integral
Now we need to solve \( \int \frac{x}{1+x^2} \, dx \). This can be done using substitution:
Let \( w = 1 + x^2 \) ⇒ \( dw = 2x \, dx \) ⇒ \( dx = \frac{dw}{2x} \).
Thus,
\[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \int \frac{1}{w} \, dw = \frac{1}{2} \ln|w| + C = \frac{1}{2} \ln(1+x^2) + C \]

### Step 5: Substitute back into the integration by parts formula
Now substituting back, we have:
\[ \int \tan^{-1}x \, dx = x \tan^{-1}x - \frac{1}{2} \ln(1+x^2) + C \]

### Step 6: Evaluate the definite integral
Now we evaluate the definite integral from \( 0 \) to \( 1 \):
\[ A = \left[ x \tan^{-1}x - \frac{1}{2} \ln(1+x^2) \right]_{0}^{1} \]

Calculating at the upper limit \( x = 1 \):
\[ = 1 \cdot \tan^{-1}(1) - \frac{1}{2} \ln(1+1^2) = \frac{\pi}{4} - \frac{1}{2} \ln(2) \]

Calculating at the lower limit \( x = 0 \):
\[ = 0 \cdot \tan^{-1}(0) - \frac{1}{2} \ln(1+0^2) = 0 - 0 = 0 \]

Thus, the area \( A \) is:
\[ A = \frac{\pi}{4} - \frac{1}{2} \ln(2) \]

### Final Answer
The area bounded by the curve \( y = \tan^{-1}x \), the X-axis, and the line \( x = 1 \) is:
\[ \frac{\pi}{4} - \frac{1}{2} \ln(2) \]