Question

The integral \(\int_{0}^{1} \cot^{-1}(1 + x + x^2) dx\) is equal to:

• A. \(2 \tan^{-1} 2 + \frac{1}{2} \log_e (\frac{5}{4}) + \frac{\pi}{2}\)
• B. \(2 \tan^{-1} 2 + \frac{1}{2} \log_e (\frac{5}{4}) - \frac{\pi}{2}\)
• C. \(2 \tan^{-1} 2 - \frac{1}{2} \log_e (\frac{5}{4}) + \frac{\pi}{2}\)
• D. \(2 \tan^{-1} 2 - \frac{1}{2} \log_e (\frac{5}{4}) - \frac{\pi}{2}\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Convert \(\cot^{-1}\) to \(\tan^{-1}\) and use the property \(\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}\) to split the integrand. Then integrate by parts.

Step 2: Key Formula or Approach:
1. \(\cot^{-1}(1 + x + x^2) = \tan^{-1}(\frac{1}{1 + x + x^2}) = \tan^{-1}(\frac{(x+1) - x}{1 + x(x+1)})\).
2. \(\int \tan^{-1} x dx = x \tan^{-1} x - \frac{1}{2} \ln(1 + x^2)\).

Step 3: Detailed Explanation:
\(I = \int_{0}^{1} [\tan^{-1}(x+1) - \tan^{-1}x] dx\).
\(I = [ (x+1) \tan^{-1}(x+1) - \frac{1}{2} \ln(1 + (x+1)^2) ]_0^1 - [ x \tan^{-1}x - \frac{1}{2} \ln(1 + x^2) ]_0^1\).
\(I = [ 2 \tan^{-1} 2 - \frac{1}{2} \ln 5 - (1 \tan^{-1} 1 - \frac{1}{2} \ln 2) ] - [ 1 \tan^{-1} 1 - \frac{1}{2} \ln 2 - 0 ]\).
\(I = 2 \tan^{-1} 2 - \frac{1}{2} \ln 5 - \frac{\pi}{4} + \frac{1}{2} \ln 2 - \frac{\pi}{4} + \frac{1}{2} \ln 2\).
\(I = 2 \tan^{-1} 2 - \frac{1}{2} \ln 5 + \ln 2 - \frac{\pi}{2} = 2 \tan^{-1} 2 - \frac{1}{2} \ln(5/4) - \frac{\pi}{2}\).

Step 4: Final Answer:
The result matches option (D).