JEE Main 2023 — Mathematics Calculus

Given,

$${\int }_{1}^{2}\frac{{t}^{4}+1}{{t}^{6}+1}dt$$∫12​t6+1t4+1​dt

$$={\int }_{1}^{2}\frac{{t}^{4}+1}{({t}^{2}+1)({t}^{4}-{t}^{2}+1)}dt$$=∫12​(t2+1)(t4−t2+1)t4+1​dt

$$[\because {a}^{3}+{b}^{3}=(a+b)({a}^{2}-ab+{b}^{2})]$$[∵a3+b3=(a+b)(a2−ab+b2)]

$$={\int }_{1}^{2}\frac{{t}^{4}+1-{t}^{2}+{t}^{2}}{({t}^{2}+1)({t}^{4}-{t}^{2}+1)}dt$$=∫12​(t2+1)(t4−t2+1)t4+1−t2+t2​dt

$$={\int }_{1}^{2}[\frac{{t}^{4}+1-{t}^{2}}{({t}^{2}+1)({t}^{4}-{t}^{2}+1)}+\frac{{t}^{2}}{({t}^{2}+1)({t}^{4}-{t}^{2}+1)}]dt$$=∫12​[(t2+1)(t4−t2+1)t4+1−t2​+(t2+1)(t4−t2+1)t2​]dt

$$={\int }_{1}^{2}[\frac{1}{({t}^{2}+1)}+\frac{{t}^{2}}{({t}^{2}+1)({t}^{4}-{t}^{2}+1)}]dt$$=∫12​[(t2+1)1​+(t2+1)(t4−t2+1)t2​]dt

$$={\int }_{1}^{2}[\frac{1}{({t}^{2}+1)}+\frac{{t}^{2}}{{t}^{6}+1}]dt$$=∫12​[(t2+1)1​+t6+1t2​]dt

$$={\int }_{1}^{2}\frac{1}{({t}^{2}+1)}dt+{\int }_{1}^{2}\frac{{t}^{2}}{{({t}^{3})}^{2}+1}dt$$=∫12​(t2+1)1​dt+∫12​(t3)2+1t2​dt

$$={[{\mathrm{tan}}^{-1}t]}_{1}^{2}+\frac{1}{3}{\int }_{1}^{2}\frac{3{t}^{2}}{{({t}^{3})}^{2}+1}dt$$=[tan−1t]12​+31​∫12​(t3)2+13t2​dt

$$=[{\mathrm{tan}}^{-1}2-{\mathrm{tan}}^{-1}1]+\frac{1}{3}{[{\mathrm{tan}}^{-1}{t}^{3}]}_{1}^{2}$$=[tan−12−tan−11]+31​[tan−1t3]12​

$$=[{\mathrm{tan}}^{-1}2-\frac{\pi }{4}]+\frac{1}{3}[{\mathrm{tan}}^{-1}8-{\mathrm{tan}}^{-1}1]$$=[tan−12−4π​]+31​[tan−18−tan−11]

$$=[{\mathrm{tan}}^{-1}2-\frac{\pi }{4}]+\frac{1}{3}[{\mathrm{tan}}^{-1}8-\frac{\pi }{4}]$$=[tan−12−4π​]+31​[tan−18−4π​]

$$={\mathrm{tan}}^{-1}2-\frac{\pi }{4}+\frac{{\mathrm{tan}}^{-1}8}{3}-\frac{\pi }{12}$$=tan−12−4π​+3tan−18​−12π​

$$={\mathrm{tan}}^{-1}2+\frac{{\mathrm{tan}}^{-1}8}{3}-\frac{\pi }{3}$$=tan−12+3tan−18​−3π​