JEE Main 2024 — 27 January, Shift 2 — Question 14

$I=\int \frac{x^{8}-x^{2}}{\left(x^{12}+3 x^{6}+1\right) \tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)} d x$

Let $\tan ^{-1}\left(\mathrm{x}^{3}+\frac{1}{\mathrm{x}^{3}}\right)=\mathrm{t}$

$\Rightarrow \frac{1}{1+\left(\mathrm{x}^{3}+\frac{1}{\mathrm{x}^{3}}\right)^{2}} \cdot\left(3 \mathrm{x}^{2}-\frac{3}{\mathrm{x}^{4}}\right) \mathrm{dx}=\mathrm{dt}$

$\Rightarrow \frac{\mathrm{x}^{6}}{\mathrm{x}^{12}+3 \mathrm{x}^{6}+1} \cdot \frac{3 \mathrm{x}^{6}-3}{\mathrm{x}^{4}} \mathrm{dx}=\mathrm{dt}$

$\mathrm{I}=\frac{1}{3} \int \frac{\mathrm{dt}}{\mathrm{t}}=\frac{1}{3} \ln |\mathrm{t}|+\mathrm{C}$

$\mathrm{I}=\frac{1}{3} \ln \left|\tan ^{-1}\left(\mathrm{x}^{3}+\frac{1}{\mathrm{x}^{3}}\right)\right|+C$

$I=\ln \left|\tan ^{-1}\left(x^{3}+\frac{1}{x^{3}}\right)\right|^{1 / 3}+C$

Hence option (1) is correct