JEE Main 2023 — Definite Integration Question with Solution

Given,

${\int }_1^2\frac{t^4+1}{t^6+1}dt$

$={\int }_1^2\frac{t^4+1}{\left(t^2+1\right)\left(t^4-t^2+1\right)}dt$

$\left[\because a^3+b^3=(a+b)(a^2-ab+b^2)\right]$

$={\int }_1^2\frac{t^4+1-t^2+t^2}{\left(t^2+1\right)\left(t^4-t^2+1\right)}dt$

$={\int }_1^2\left[\frac{t^4+1-t^2}{\left(t^2+1\right)\left(t^4-t^2+1\right)}+\frac{t^2}{\left(t^2+1\right)\left(t^4-t^2+1\right)}\right]dt$

$={\int }_1^2\left[\frac{1}{\left(t^2+1\right)}+\frac{t^2}{\left(t^2+1\right)\left(t^4-t^2+1\right)}\right]dt$

$={\int }_1^2\left[\frac{1}{\left(t^2+1\right)}+\frac{t^2}{t^6+1}\right]dt$

$={\int }_1^2\frac{1}{\left(t^2+1\right)}dt+{\int }_1^2\frac{t^2}{{\left(t^3\right)}^2+1}dt$

$={\left[{\tan }^{-1}t\right]}_1^2+\frac{1}{3}{\int }_1^2\frac{3t^2}{{\left(t^3\right)}^2+1}dt$

$=\left[{\tan }^{-1}2-{\tan }^{-1}1\right]+\frac{1}{3}{\left[{\tan }^{-1}{t}^3\right]}_1^2$

$=\left[{\tan }^{-1}2-\frac{\mathrm{\pi }}{4}\right]+\frac{1}{3}\left[{\tan }^{-1}8-{\tan }^{-1}1\right]$

$=\left[{\tan }^{-1}2-\frac{\mathrm{\pi }}{4}\right]+\frac{1}{3}\left[{\tan }^{-1}8-\frac{\mathrm{\pi }}{4}\right]$

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

$={\tan }^{-1}2+\frac{{\tan }^{-1}8}{3}-\frac{\mathrm{\pi }}{3}$