JEE Main 2026 — 4 April, Evening Shift — Question 44

The given integral is:

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

Simplify the Integrand Using the identity $\cot^{-1} \theta = \tan^{-1} \left( \frac{1}{\theta} \right)$:

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

Rearranging the argument to fit the identity $\tan^{-1} \left( \frac{A-B}{1+AB} \right) = \tan^{-1} A - \tan^{-1} B$:

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

Definite Integration

$$I = \int_{0}^{1} \tan^{-1}(x+1) \, dx - \int_{0}^{1} \tan^{-1}x \, dx$$

Recall the general integral formula: $\int \tan^{-1} u \, du = u \tan^{-1} u - \frac{1}{2} \ln(1+u^2)$.

Evaluating the first part:

$$\left[ (x+1) \tan^{-1}(x+1) - \frac{1}{2} \ln(1+(x+1)^2) \right]_0^1$$ $$= \left( 2 \tan^{-1} 2 - \frac{1}{2} \ln 5 \right) - \left( 1 \tan^{-1} 1 - \frac{1}{2} \ln 1 \right)$$ $$= 2 \tan^{-1} 2 - \frac{1}{2} \ln 5 - \frac{\pi}{4}$$

Evaluating the second part:

$$\left[ x \tan^{-1} x - \frac{1}{2} \ln(1+x^2) \right]_0^1$$ $$= \left( 1 \tan^{-1} 1 - \frac{1}{2} \ln 2 \right) - (0)$$ $$= \frac{\pi}{4} - \frac{1}{2} \ln 2$$

Combine and Simplify

$$I = \left( 2 \tan^{-1} 2 - \frac{1}{2} \ln 5 - \frac{\pi}{4} \right) - \left( \frac{\pi}{4} - \frac{1}{2} \ln 2 \right)$$ $$I = 2 \tan^{-1} 2 - \frac{1}{2} \ln 5 + \frac{1}{2} \ln 2 - \frac{\pi}{2}$$

$I=2\tan^{-1}2 - \frac{1}{2}\log_e\left(\frac{5}{2}\right) -\frac{\pi}{2}$