Question

Evaluate the integral
\[ \int_{-1}^1 \left( \tan^{-1} \left( \frac{x}{x+1} \right) + \tan^{-1} \left( \frac{x+1}{x} \right) \right) \, dx \]

• A. \( \frac{\pi}{2} \)
• B. \( \pi \)
• C. \( \frac{3\pi}{2} \)
• D. \( 2\pi \)

Hint:

When encountering sums of inverse tangent functions, apply the identity for the sum of tangents to simplify the expression before integration.

Answer 1

The Correct Option is B

Step 1: Use the identity for inverse tangent addition.
The given expression involves a sum of two inverse tangent functions. We can apply the following identity for the sum of inverse tangents:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
We will apply this identity to simplify the sum of \( \tan^{-1} \left( \frac{x}{x+1} \right) \) and \( \tan^{-1} \left( \frac{x+1}{x} \right) \).

Step 2: Apply the identity.
Let \( a = \frac{x}{x+1} \) and \( b = \frac{x+1}{x} \). Then,
\[ \tan^{-1} \left( \frac{x}{x+1} \right) + \tan^{-1} \left( \frac{x+1}{x} \right) = \tan^{-1} \left( \frac{\frac{x}{x+1} + \frac{x+1}{x}}{1 - \frac{x}{x+1} \cdot \frac{x+1}{x}} \right) \]

Step 3: Simplify the numerator.
Simplify the numerator of the fraction inside the inverse tangent:
\[ \frac{x}{x+1} + \frac{x+1}{x} = \frac{x^2 + (x+1)^2}{x(x+1)} = \frac{x^2 + x^2 + 2x + 1}{x(x+1)} = \frac{2x^2 + 2x + 1}{x(x+1)} \]

Step 4: Simplify the denominator.
Now simplify the denominator:
\[ 1 - \frac{x}{x+1} \cdot \frac{x+1}{x} = 1 - 1 = 0 \]

Step 5: Analyze the result.
So the expression inside the inverse tangent function simplifies to infinity. Therefore, the sum of the two inverse tangents is equal to \( \frac{\pi}{2} \).

Step 6: Integrate the result.
Now, we need to integrate the result over the limits from \( -1 \) to \( 1 \). Since the result is \( \pi \), the integral becomes:
\[ \int_{-1}^1 \pi \, dx = \pi \times \left[ x \right]_{-1}^1 = \pi \times (1 - (-1)) = 2\pi \]
Thus, the value of the integral is \( \boxed{\pi} \).