JEE Main 2026 Apr 4 Shift 2, Mathematics Q20: Integration by Parts

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

So $I={\int }_0^1{\tan }^{-1}(x+1)dx-{\int }_0^1{\tan }^{-1}xdx$.

Standard result: $\int {\tan }^{-1}udu=u{\tan }^{-1}u-\frac{1}{2}\ln (1+u^2)+C$.

${\int }_0^1{\tan }^{-1}(x+1)dx$ — let $u=x+1$, runs from 1 to 2:

$={\left[u{\tan }^{-1}u-\frac{1}{2}\ln (1+u^2)\right]}_1^2=\left(2{\tan }^{-1}2-\frac{1}{2}\ln 5\right)-\left(\frac{\pi }{4}-\frac{1}{2}\ln 2\right)$.

${\int }_0^1{\tan }^{-1}xdx=\frac{\pi }{4}-\frac{1}{2}\ln 2$.

Subtract: $I=2{\tan }^{-1}2-\frac{1}{2}\ln 5-\frac{\pi }{4}+\frac{1}{2}\ln 2-\frac{\pi }{4}+\frac{1}{2}\ln 2$.

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