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

(D) $2 \tan ^{-1} 2-\frac{1}{2} \log _{\mathrm{e}}\left(\frac{5}{4}\right)-\frac{\pi}{2}$

(A) $2 \tan ^{-1} 2+\frac{1}{2} \log _{\mathrm{e}}\left(\frac{5}{4}\right)+\frac{\pi}{2}$

(B) $2 \tan ^{-1} 2+\frac{1}{2} \log _{e}\left(\frac{5}{4}\right)-\frac{\pi}{2}$

(C) $2 \tan ^{-1} 2-\frac{1}{2} \log _{e}\left(\frac{5}{4}\right)+\frac{\pi}{2}$

Definite Integrals — JEE Main · 2026 · 04 Apr 2026 · Shift 2 Question with Solution

The integral

\int_{0}^{1} cot^{- 1} (1 + x + x^{2}) dx

is equal to :

(A) $2 tan^{- 1} 2 + \frac{1}{2} lo g_{e} (\frac{5}{4}) + \frac{π}{2}$
(B) $2 tan^{- 1} 2 + \frac{1}{2} lo g_{e} (\frac{5}{4}) - \frac{π}{2}$
(C) $2 tan^{- 1} 2 - \frac{1}{2} lo g_{e} (\frac{5}{4}) + \frac{π}{2}$
(D) $2 tan^{- 1} 2 - \frac{1}{2} lo g_{e} (\frac{5}{4}) - \frac{π}{2}$ ✓ Correct

Answer: (D)

Since

cot^{- 1} (1 + x + x^{2}) = tan^{- 1} \frac{1}{1 + x + x ^{2}} = tan^{- 1} \frac{( x + 1 ) - x}{1 + ( x + 1 ) x} = tan^{- 1} (x + 1) - tan^{- 1} x

, the integral is

\int_{0}^{1} [tan^{- 1} (x + 1) - tan^{- 1} x] d x

. Integrating by parts,

\int tan^{- 1} x d x = x tan^{- 1} x - \frac{1}{2} ln (1 + x^{2})

and

\int tan^{- 1} (x + 1) d x = x tan^{- 1} (x + 1) - \frac{1}{2} ln (1 + (x + 1)^{2}) + tan^{- 1} (x + 1)

. Evaluating from

0

1

gives

(2 tan^{- 1} 2 - \frac{1}{2} ln \frac{5}{2} - \frac{π}{4}) - (\frac{π}{4} - \frac{1}{2} ln 2)

. Combining the logarithms,

\frac{1}{2} ln \frac{5}{2} - \frac{1}{2} ln 2 = \frac{1}{2} ln \frac{5}{4}

, and the integral equals

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

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option