The integral is equal to: [2026]

Question

The integral $\int_{0}^{1} \cot^{- 1} (1 + x + x^{2}) dx$ is equal to: [2026]

Smart Achievers · Accepted Answer

(4)

Let $I = \int_{0}^{1} \cot^{- 1} (1 + x + x^{2}) dx$

$= \int_{0}^{1} \tan^{- 1} (\frac{1}{1 + x + x^{2}}) d x$

$= \int_{0}^{1} \tan^{- 1} (\frac{(x + 1) - x}{1 + x (x + 1)}) d x$

$= \int_{0}^{1} (\tan^{- 1} (x + 1) - \tan^{- 1} x) d x$

$= \int_{0}^{1} \tan^{- 1} (x + 1) d x - \int_{0}^{1} \tan^{- 1} x d x$

$= \int_{0}^{2} \tan^{- 1} u d u - \int_{0}^{1} \tan^{- 1} x d x$ [Put x + 1 = u in 1st integration]

$= \int_{0}^{2} \tan^{- 1} x d x - \int_{0}^{1} \tan^{- 1} x d x$

$= {[x \tan^{- 1} x]}_{1}^{2} - {[\frac{1}{2} \ln (1 + x^{2})]}_{1}^{2} - {[x \tan^{- 1} x]}_{0}^{1} + {[\frac{1}{2} \ln (1 + x^{2})]}_{0}^{1}$

$= 2 \tan^{- 1} 2 - \tan^{- 1} 1 - [\frac{1}{2} \ln 5 - \frac{1}{2} \ln 2] - \tan^{- 1} 1 + \frac{1}{2} \ln 2$

$= 2 \tan^{- 1} 2 - 2 \tan^{- 1} 1 - \frac{1}{2} \ln 5 + \ln 2$

$= 2 \tan^{- 1} 2 - 2 \times \frac{π}{4} + \ln 5$

$= 2 \tan^{- 1} 2 - \frac{π}{2} - \frac{1}{2} \ln \frac{5}{4}$

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