Definite Integration
General
Grade 12
Question:
<div>$\int_{0}^{1} \cot^{-1}(1-x+x^2) dx$ equals -</div>
<div>$\frac{\pi}{2} + \log 2$</div>
<div>$\frac{\pi}{2} - \log 2$</div>
<div>$\pi - \log 2$</div>
<div>none of these</div>
Step-by-Step Solution
Key Concept: General
<div>$I = \int_{0}^{1} \tan^{-1}\left(\frac{1}{1-x+x^2}\right) dx = \int_{0}^{1} \tan^{-1}\left(\frac{x+(1-x)}{1-x(1-x)}\right) dx = \int_{0}^{1} [\tan^{-1} x + \tan^{-1}(1-x)] \ dx = \int_{0}^{1} \tan^{-1} x \ dx + \int_{0}^{1} \tan^{-1}(1-x) \ dx = 2 \int_{0}^{1} \tan^{-1} x \ dx = 2 \left[ x \tan^{-1} x - \frac{1}{2} \log(1+x^2) \right]_{0}^{1} = 2 \frac{\pi}{4} - \log 2 = \frac{\pi}{2} - \log 2$</div>
Correct Answer: B