JEE Main 2026 — Mathematics Calculus

$$I = \int_{0}^{1}\cot^{-1}(1+x+x^2)dx$$I=∫01​cot−1(1+x+x2)dx

$$I = \int_{0}^{1}\tan^{-1}\left(\dfrac{1}{1+x+x^2}\right)dx$$I=∫01​tan−1(1+x+x21​)dx

$$I = \int_{0}^{1}\tan^{-1}\left(\dfrac{(x+1)-x}{1+(x+1)x}\right)dx$$I=∫01​tan−1(1+(x+1)x(x+1)−x​)dx

$$I = \int_{0}^{1}(\tan^{-1}(x+1)-\tan^{-1}x)dx$$I=∫01​(tan−1(x+1)−tan−1x)dx

$$I = \int_{0}^{1}\tan^{-1}(x+1)dx - \int_{0}^{1}\tan^{-1}x dx$$I=∫01​tan−1(x+1)dx−∫01​tan−1xdx

Substituting $x+1 = t$x+1=t in the first integral:

$$I = \int_{1}^{2}\tan^{-1}x dx - \int_{0}^{1}\tan^{-1}x dx$$I=∫12​tan−1xdx−∫01​tan−1xdx

Using integration by parts, $\displaystyle \int \tan^{-1}x dx = x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)$∫tan−1xdx=xtan−1x−21​loge​(1+x2)

$$I = \left[x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)\right]_{1}^{2} - \left[x\tan^{-1}x - \dfrac{1}{2}\log_e(1+x^2)\right]_{0}^{1}$$I=[xtan−1x−21​loge​(1+x2)]12​−[xtan−1x−21​loge​(1+x2)]01​

$$I = \left(2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 - \left(\dfrac{\pi}{4} - \dfrac{1}{2}\log_e 2\right)\right) - \left(\dfrac{\pi}{4} - \dfrac{1}{2}\log_e 2 - 0\right)$$I=(2tan−12−21​loge​5−(4π​−21​loge​2))−(4π​−21​loge​2−0)

$$I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 - \dfrac{\pi}{2} + \log_e 2$$I=2tan−12−21​loge​5−2π​+loge​2

$$I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e 5 + \dfrac{1}{2}\log_e 4 - \dfrac{\pi}{2}$$I=2tan−12−21​loge​5+21​loge​4−2π​

$$I = 2\tan^{-1}2 - \dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right) - \dfrac{\pi}{2}$$I=2tan−12−21​loge​(45​)−2π​

Answer: $2\tan^{-1}2-\dfrac{1}{2}\log_e\left(\dfrac{5}{4}\right)-\dfrac{\pi}{2}$2tan−12−21​loge​(45​)−2π​