Find the anti-derivative and $$ \left.\begin{array}{l}\text { simplify. } \ =\frac{1}{x \sqrt{x^{6}-1}} d x \ {[?]} \ \sec ^{-1} \mid x\end{array} \right\rvert\,+C $$ Hint: Let $$ u=x^{3} $$

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We wish to compute
$$
\int \frac{1}{x\sqrt{x^6-1}}\,dx.
$$
To begin, let
$$
u=x^3.
$$
Then
$$
du = 3x^2\,dx\quad\Rightarrow\quad dx=\frac{du}{3x^2}.
$$

Note that since $$x^6=(x^3)^2=u^2$$, we have
$$
\sqrt{x^6-1}=\sqrt{u^2-1}.
$$
Also, using the substitution $$u=x^3$$, we observe that
$$
x^3=u.
$$

Now, substitute into the original integral:
$$
\int \frac{1}{x\sqrt{u^2-1}}\,dx = \int \frac{1}{x\sqrt{u^2-1}}\cdot\frac{du}{3x^2} 
=\frac{1}{3}\int \frac{du}{x^3\sqrt{u^2-1}}.
$$
Since $$x^3=u$$, this simplifies to
$$
\frac{1}{3}\int \frac{du}{u\sqrt{u^2-1}}.
$$

Recall that for $$u\ge1$$, the derivative of the inverse secant function is
$$
\frac{d}{du}\left(\sec^{-1}(u)\right)=\frac{1}{|u|\sqrt{u^2-1}},
$$
and when $$u>0$$ we have $$|u|=u$$. Thus,
$$
\int \frac{du}{u\sqrt{u^2-1}} = \sec^{-1}(u)+C.
$$

Therefore, the antiderivative becomes
$$
\frac{1}{3}\sec^{-1}(u)+C.
$$
Substitute back $$u=x^3$$ to obtain the final answer:
$$
\frac{1}{3}\sec^{-1}(x^3)+C.
$$
The anti-derivative is $$ \frac{1}{3} \sec^{-1}(x^3) + C $$.

Find the anti-derivative and

Hint: Let

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Find the anti-derivative and Hint: Let