JEE Mains · Maths · STD 12 - 7.2 definite integral
The integral \(\int_{0}^{1}\cot^{-1}(1+x+x^2)dx\) is equal to:
- A \(2\tan^{-1}2+\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)+\dfrac{\pi}{2}\)
- B \(2\tan^{-1}2+\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}\)
- C \(2\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)+\dfrac{\pi}{2}\)
- D \(2\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}\)
Answer & Solution
Correct Answer
(D) \(2\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
\(I = \int_{0}^{1}\cot^{-1}(1+x+x^2)dx\) \(I = \int_{0}^{1}\tan^{-1}\left(\dfrac{1}{1+x+x^2}\right)dx\) \(I = \int_{0}^{1}\tan^{-1}\left(\dfrac{(x+1)-x}{1+(x+1)x}\right)dx\) \(I = \int_{0}^{1}(\tan^{-1}(x+1)-\tan^{-1}x)dx\)…
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