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Step-by-Step Solution
Step 1: Express the integral in a manageable form
The given integral is
$ \int \frac{\bigl(x^8 - x^2\bigr)\,dx}{\bigl(x^{12} + 3x^6 + 1\bigr)\,\tan^{-1}\Bigl(x^3 + \frac{1}{x^3}\Bigr)}. $
First, notice that we can factor the numerator to make it simpler:
$ x^8 - x^2 = x^2 \bigl(x^6 - 1\bigr). $
Step 2: Identify a useful substitution involving $x^3 + \frac{1}{x^3}$
Let
$ u = x^3 + \frac{1}{x^3}. $
Then observe:
$ x^6 + \frac{1}{x^6} = u^2 - 2, $
and hence
$ x^{12} + 3x^6 + 1 = x^6 \bigl(x^6 + \frac{1}{x^6} + 3 \bigr) = x^6 (u^2 + 1). $
So the integral becomes
$ \int \frac{x^2 \bigl(x^6 - 1\bigr)}{x^6 \,\bigl(u^2 + 1\bigr)\,\tan^{-1}(u)} \, dx
= \int \frac{x^6 - 1}{x^4} \,\frac{1}{\bigl(u^2 + 1\bigr)\,\tan^{-1}(u)} \, dx. $
Step 3: Compute $du$ and rewrite the integrand
Differentiate
$ u = x^3 + \tfrac{1}{x^3} $
with respect to $x$:
$ \frac{du}{dx} = 3x^2 - \frac{3}{x^4}
= 3\Bigl(x^2 - \frac{1}{x^4}\Bigr)
= 3\, \frac{x^6 - 1}{x^4}. $
Therefore,
$ \frac{x^6 - 1}{x^4} \, dx = \frac{1}{3}\,du. $
Substituting back into the integral:
$ \int \frac{x^6 - 1}{x^4}\,\frac{1}{(u^2 + 1)\,\tan^{-1}(u)}\,dx
= \int \frac{1}{(u^2 + 1)\,\tan^{-1}(u)} \,\frac{du}{3}
= \frac{1}{3} \int \frac{du}{(u^2 + 1)\,\tan^{-1}(u)}. $
Step 4: Second substitution to handle $\tan^{-1}(u)$
Let
$ v = \tan^{-1}(u). $
Then
$ dv = \frac{1}{1 + u^2} \, du,\; \text{hence}\; du = (1 + u^2)\,dv. $
Substituting into the integral gives:
$ \int \frac{1}{(u^2 + 1)\,\tan^{-1}(u)} \, du
= \int \frac{1}{v} \,\frac{1}{u^2+1} \,(u^2 + 1)\,dv
= \int \frac{1}{v}\, dv
= \ln|v| + C. $
Since $v = \tan^{-1}(u),$ we have
$ \ln|\tan^{-1}(u)| + C. $
Step 5: Incorporate the constant factor and revert to $x$
Recall the factor $\tfrac{1}{3}$ outside the integral. Thus, the solution becomes:
$ \frac{1}{3}\,\ln \Bigl|\tan^{-1}\bigl(u\bigr)\Bigr| + C
= \frac{1}{3}\,\ln \Bigl|\tan^{-1}\Bigl(x^3 + \frac{1}{x^3}\Bigr)\Bigr| + C. $
We can write
$ \frac{1}{3}\,\ln|z| = \ln\bigl(|z|^{\tfrac{1}{3}}\bigr). $
Therefore, the final answer is:
$ \boxed{
\ln \Bigl|\tan^{-1}\Bigl(x^3 + \frac{1}{x^3}\Bigr)\Bigr|^{\tfrac{1}{3}} + C
}. $
