Simple Integral

$\begin{aligned}3F(x)= \int\dfrac{3x^2\mathrm dx}{\underbrace{x^3}_{\color{#D61F06}t}\sqrt{x^6-1}}=&\int\dfrac{\mathrm{d}t}{\underbrace{t}_{\color{#D61F06}\sec y}\sqrt{t^2-1}}\\=&\int\dfrac{\cancel{\sec y\tan y}\mathrm{d}y}{\cancel{\sec y\sqrt{\sec^2 y-1}}}\\=& y+C\\=&\sec^{-1} x^3+C\end{aligned}$

$\boxed{\implies F(x)=\dfrac{\sec^{-1} x^3}{3}}~~,F(1)=0$

$F(n)=\dfrac{\sec^{-1} n^3}{3}=\dfrac{\pi}{9}$ $\implies n^3=\sec \dfrac{\pi}{3}=\boxed{\color{#20A900}2}$

$F(x) = \displaystyle{\int \dfrac{x^5 dx}{x^6 \sqrt{x^6 - 1}}}$

Now, let $u = x^6 \rightarrow du = 6x^5$ . We have $\displaystyle{\int \dfrac{x^5 dx}{x^6 \sqrt{x^6 - 1}} = \dfrac{1}{6}\int \dfrac{du}{u\sqrt{u-1}}}$

Letting $z = \sqrt{u-1} \rightarrow u = z^2 + 1 \rightarrow du = 2zdz$

$\begin{aligned} \int \dfrac{x^5 dx}{x^6 \sqrt{x^6 - 1}} &=& \dfrac{1}{6}\int \dfrac{du}{u\sqrt{u-1}}\\ &=& \dfrac{1}{6}\int \dfrac{2z \, dz}{(z^2 + 1)z}\\ &=& \dfrac{1}{3} \int \dfrac{dz}{z^2 + 1}\\ &=& \dfrac{1}{3} \tan^{-1} z + C\\ &=& \dfrac{1}{3} \tan^{-1} \sqrt{x^6 - 1} + C \end{aligned}$

Now, $F(1) = 0$ gives $C = 0$ meaning $F(x) = \dfrac{1}{3} \tan^{-1} \sqrt{x^6 - 1}$ .

If we troubleshoot the problem, I think it should've been asked "Determine the value of $|n^3|$ for which $F(n) = \dfrac{\pi}{9}$ ". If that was the problem, then

$\begin{aligned} \dfrac{1}{3} \tan^{-1} \sqrt{n^6 - 1} &=& \dfrac{\pi}{9}\\ n^6 - 1 &=& 3\\ |n^3| &=& \boxed{2} \end{aligned}$