JEE 2023 — Mathematics PYQ

Solution:

$\frac{t^4+1}{t^6+1}=\frac{(t^4-t^2+1)+t^2}{(t^2+1)(t^4-t^2+1)}=\frac{1}{t^2+1}+\frac{t^2}{t^6+1}$.

$\int \frac{1}{t^2+1}dt={\tan }^{-1}t$.

$\int \frac{t^2}{t^6+1}dt$ (Put $t^3=u$) $=\frac{1}{3}{\tan }^{-1}(t^3)$.

Limits $(1\to 2)$: $({\tan }^{-1}2-{\tan }^{-1}1)+\frac{1}{3}({\tan }^{-1}8-{\tan }^{-1}1)$.

$={\tan }^{-1}2+\frac{1}{3}{\tan }^{-1}8-\frac{4}{3}(\frac{\pi }{4})={\tan }^{-1}2+\frac{1}{3}{\tan }^{-1}8-\frac{\pi }{3}$.

• Correct Option: (4).