Mathematics - Indefinite Integration Question with Solution | TestHub

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

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

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

$\Rightarrow \frac{1}{1+\left({\mathrm{x}}^6+{\displaystyle \frac{1}{x^6}}+2\right)}·\left(3{\mathrm{x}}^2-\frac{3}{{\mathrm{x}}^4}\right)\mathrm{dx}=\mathrm{dt}$

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

$\Rightarrow \frac{\left({\mathrm{x}}^8-{\mathrm{x}}^2\right)}{{\mathrm{x}}^{12}+3{\mathrm{x}}^6+1}\mathrm{dx}=\frac{\mathrm{dt}}{3}$

$\Rightarrow \mathrm{I}=\frac{1}{3}\int \frac{1}{\mathrm{t}}\mathrm{dt}$

$\Rightarrow \mathrm{I}=\frac{1}{3}\log \left|\mathrm{t}\right|+\mathrm{C}$

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

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

Hence optin $\left(\mathrm{A}\right)$ is correct