The challenge is to integrate
$$ \int [(x^2 - 1)(x + 1)]^{(-2/3)} dx $$
using $u = tan^{(-1)} x,~$ minimizing the artificiality involved.
This is the closest I could come.
$\displaystyle [E_1]:~ \forall ~k ~\in ~\mathbb{Z} ~ \text{such that} ~ k \neq 0$
$$
\frac{d}{du} \left(\frac{\sin u - \cos u}{\sin u + \cos u}\right)^k
~= k \left[\frac{\sin u - \cos u}{\sin u + \cos u}\right]^{(k-1)} \times
\frac{2}{\left(\sin u + \cos u\right)^2}.$$
I discovered $[E_1]$, which may be routinely verified, simply by using
straight forward methods to simplify the integrand in terms of $u$, reverse
engineering the form that the solution would need to take, and then experimenting.
Personally, I know of no reasonable way of independently deciding that $[E_1]$
will be important. Nor do I know of any reasonable way to legitimately
employ integration by parts (for example). Anyway...
Setting $u = \tan^{(-1)} x$ gives
(1) $~x = \tan u$
(2) $~dx = \frac{1}{\cos^2 u} du.$
(3) $~(x+1) = (1 + \tan u) ~=~ \frac{\cos u + \sin u}{\cos u}.$
(4) $~(x^2 - 1) = (\tan^2 u - 1) ~=~ \frac{\sin^2 u - \cos^2 u}{\cos^2 u}.$
We have
$$ J ~=~ \int \left[\left(\tan^2 u - 1\right)\left(1 + \tan u\right)\right]^{(-2/3)}
\times \frac{1}{\cos^2 u} ~du $$
$$ =~ \int \left[\tan^2 u - 1 + \tan^3 u - \tan u\right]^{(-2/3)}
\times \frac{1}{\cos^2 u} ~du $$
$$ =~ \int \left[\frac{\sin^2 u \cos u - \cos^3 u + \sin^3 u - \sin u \cos^2 u}{\cos^3 u}\right]^{(-2/3)}
\times \frac{1}{\cos^2 u} ~du $$
$$ =~ \int \left[\sin^2 u \cos u - \cos^3 u + \sin^3 u - \sin u \cos^2 u\right]^{(-2/3)}
~du $$
$$ =~ \int \left[\left(\sin^2 u - \cos^2 u\right)\left(\sin u + \cos u\right)\right]^{(-2/3)}
~du. $$
$$ J ~=~ \int \left[\left(\sin u - \cos u\right)\left(\sin u + \cos u\right)^2\right]^{(-2/3)}
~du. $$
At this point, with $E_1$ established, I used reverse-engineering on the answer:
$$ I ~=~ \frac{3}{2} \times \left[\frac{\sin u - \cos u}{\sin u + \cos u}\right]^{(1/3)}.$$
Using $I$ as a goal, using $[E_1]$ as a model, and setting $k = \frac{1}{3}$
allowed $J$ to be re-expressed as
$$\left(\frac{3}{2}\right) \left(\frac{1}{3}\right)
~\int \left[\frac{\left(\sin u - \cos u\right)}
{\left(\sin u + \cos u\right)}\right]^{(-2/3)} \times \frac{2}{\left(\sin u + \cos u\right)^2} du.$$
In accordance with $[E_1]$, this results in $\int J ~=~ I.$
Addendum
My original answer involved reverse-engineering of the known answer to prompt exploration of the equation $[E_1]$. As stated in the answer, I considered this an artificial contrivance. This Addendum is intended to provide a more satisfactory completion of the problem.
Part of my original answer involved applying the required $u$-substitution and then converting the indefinite integral into
$$ J = \int \left[\left(\sin u - \cos u\right)\left(\sin u + \cos u\right)^2\right]^{(-2/3)} ~du. $$
I consider that portion of my original answer acceptable, so the problem is reduced to integrating $J$.
$\underline{\text{General Considerations}}$
In attempting to integrate $J$, I considered the following issues:
Although this particular problem is known to have a closed form solution, in general, attempting to integrate problems of this type may be a fool's errand; there may not be a closed form solution. I know of no mathematics, at the undergraduate level, that allows one to quickly determine whether such a solution exists.
Can Taylor polynomials be used to attack the problem? The idea is to convert the sine and cosine functions into their Taylor polynomials, use this conversion to compute the entire integrand of $J$ into a single polynomial of infinite degree, study this resultant
polynomial, and then (somehow) convert this resultant polynomial back into a closed form expression involving (for example) the sine and cosine functions.
I have never seen this approach attempted.
Given the $(-2/3)$ exponent that occurs in the integrand of $J$, the following method suggests itself:
Let $f(u) = \left[\left(\sin u - \cos u\right)\left(\sin u + \cos u\right)^2\right].$
Search for an appropriate function $g(u)$ and convert the integrand of $J$ into
$\left[f(u) \times g(u)\right]^{(-2/3)}
\times \left[g(u)\right]^{(2/3)}.$
The sole constraint on $g(u)$ is that
$\displaystyle \frac{d}{du} \left[f(u) \times g(u)\right] ~\text{must} ~=~ \left[g(u)\right]^{(2/3)}.$
Then $J$ will equal $3 \times \left[f(u) \times g(u)\right]^{(1/3)}.$
Personally, I see nothing illegitimate about this approach, as it is merely reacting to the $(-2/3)$ exponent which occurs in the integrand of $J$.
Suppose that you are given that there is a closed form solution for $J$, and you notice the $(-2/3)$ exponent that appears in the integrand of $J$. This doesn't mean that the approach suggested in the previous note is the only viable integration strategy. In fact, an argument could be made that ywrht_'s answer to the original query is a case in point.
Not only did the approach taken in ywrht_'s answer never occur to me, but it is plausible that with a new but similar problem, (again) no special insight would occur to me. The approach indicated in the previous note, which is the approach that will be taken in this Addendum, dispenses with any attempt at elegance, and attacks the problem with brute force.
$\underline{\text{My Work}}$
Let $~R(u) \equiv (\sin u - \cos u),~$
and let $~S(u) \equiv (\sin u + \cos u).$
Then, $~R'(u) = S(u), ~S'(u) = -R(u),~$ and
$~f(u) = \left\{R(u) \times \left[S(u)\right]^2 \right\}.$
Also, $[R(u)]^2 + [S(u)]^2 = 2$, for any value of $u$.
To make the remainder of this Addendum more readable, $R$ will be used to represent $R(u)$ and $S$ will be used to represent $S(u).$
With $f(u) = RS^2,$
$$f'(u) = \left[R \times (-2RS)\right] ~+~
\left[S^2 \times (S)\right] ~=~ (S^3 - 2R^2S).$$
Given the constraint that will be imposed on $g(u)$, and given the nature of $f(u)$, my best try in the search for $g(u)$ is
$$ g(u) = D^3 \times R^{(3A)} \times S^{(3B)}$$
where $D \neq 0, A,$ and $B$ are real numbers to be determined.
This means that the constraint can be expressed as
$$\frac{d}{du}\left[f(u) \times g(u)\right]
~=~ D^2 \times R^{(2A)} \times S^{(2B)}.$$
Note:
See the Addendum's "Legitimacy of Approach" section for consideration of the specification of $g(u).$
Anyway, with $g(u)$ so specified, as a guess, the remainder of this Addendum will analytically derive the unique set of parameters for $D, A,$ and $B$, that generate a satisfying function $g(u)$.
$$g'(u) ~=~ D^3 \times
\left\{
\left[R^{(3A)} \times 3BS^{(3B-1)} \times (-R)\right] ~+~
\left[S^{(3B)} \times 3AR^{(3A-1)} \times (S)\right]
\right\}
$$
$$=~ D^3 \times
\left[3AR^{(3A-1)} S^{(3B+1)} ~-~
3BR^{(3A+1)} S^{(3B-1)}
\right]
$$
Now, $~\displaystyle \frac{d}{du}
\left[f(u) \times g(u)\right]~$ can be computed as
$\displaystyle =~ RS^2 \times D^3
\left[3AR^{(3A-1)} S^{(3B+1)} ~-~ 3BR^{(3A+1)} S^{(3B-1)}
\right]
$
$\displaystyle +~ D^3 R^{(3A)}S^{(3B)} \times \left[S^3 - 2R^2S\right] $
$\displaystyle =~ D^3 \times
\left[
(3A + 1)R^{(3A)}S^{(3B+3)} ~-~ (3B + 2)R^{(3A+2)}S^{(3B+1)}
\right]
$
The next
section of the Addendum computes the parameters, $D, A,$ and $B$.
$\underline{\text{Determination of Parameters}}$
The parameters of $D,A,$ and $B$ must be computed to
satisfy the constraint that
$$\frac{d}{du} \left[f(u) \times g(u)\right]
~=~ g(u)^{(2/3)}.$$
The only tool available is that $R^2 + S^2$ = 2. The constraint is:
$\displaystyle ~D^3 \times
\left[
(3A + 1)R^{(3A)}S^{(3B+3)} ~-~ (3B+2)R^{(3A+2)}S^{(3B+1)}
\right]
$
$\displaystyle =~ D^2 \times R^{(2A)} \times S^{(2B)}.$
Since the RHS of the constraint above is a common factor of each term in the LHS, the constraint can be reduced to
$[F_1]$
$\displaystyle D \times
\left[
(3A + 1)R^{(A)}S^{(B+3)} ~-~ (3B+2)R^{(A+2)}S^{(B+1)}
\right] ~=~ 1$.
Although the eventual solution can be derived by looking for common factors and applying guesswork, no guesswork is actually necessary. Keeping in mind that $R$ symbolizes $R(u)$, and $S$ symbolizes $S(u)$, the above equation indicates that the LHS is constant, for all values of $u$.
Therefore, the parameters for $D, A,$ and $B$ can be derived by requiring that the first derivative of the LHS = 0. This gives:
$\displaystyle ~~
(3A+1)\left[R^{(A)}(B+3)S^{(B+2)}(-R) + S^{(B+3)}(A)R^{(A-1)}(S)
\right]$
$\displaystyle -~
(3B+2)\left[R^{(A+2)}(B+1)S^{(B)}(-R) + S^{(B+1)}(A+2)R^{(A+1)}(S)
\right]$
$\displaystyle =~ 0.$
Dividing by the common factors of
$~\left[R^{(A-1)} \times S^{(B)}\right]~$
and collecting terms, this gives
$\displaystyle ~~
(3A+1)\left[R^{(2)}(B+3)S^{(2)}(-1) + S^{(4)}(A)
\right]$
$\displaystyle -~
(3B+2)\left[R^{4}(B+1)(-1) + S^{(2)}(A+2)R^{(2)}
\right]$
$\displaystyle =~ 0 ~~\Rightarrow $
$[F_2]$
$\displaystyle R^4\left[(3B+2)(B+1)\right] ~+~
R^2 S^2 (-1)\left[(3A+1)(B+3) + (3B+2)(A+2)\right]$
$\displaystyle ~+~ S^4 \left[(3A+1)(A)\right] ~=~ 0.$
At this point, the analysis may be validly shortcircuited by remembering
that $R$ and $S$ are each symbolizing non-constant functions of $u$, and
that $R^2 + S^2 = 2.$
Note that $S^2 = [S(u)]^2 = (\sin u + \cos u)^2 = (1 + \sin 2u).$
Let $T = S^2$ and suppose that real coefficients $c_1, c_2, c_3$
were found such that
$[F_3]$
$c_1T^2 + c_2T + c_3 = 0,~$ which is a constant.
Taking the derivative of $[F_3]$, this would imply that
$2Tc_1 + c_2 = 0 ~\Rightarrow $
[since $~T = (1 + \sin 2u)~$ which varies with $u$] $~c_1 = 0 \Rightarrow$
$c_2 = 0 \Rightarrow c_3 = 0.$
Similarly consider an equation
$[F_4]$
$k_1 R^4 + k_2 R^2 S^2 + k_3 S^4 = 0.$
Again setting $T = S^2,$ and substituting $(2 - T)$ for $R^2$ gives
$k_1 (2-T)^2 + k_2 (2-T)(T) + k_3 (T^2) = 0 ~\Rightarrow $
$k_1 (4 - 4T + T^2) + k_2 (2T - T^2) + k_3 (T^2) = 0 \Rightarrow $
$T^2(k_1 - k_2 + k_3) + T(- 4k_1 + 2k_2) + (4)(k_1) = 0.$
Based on the previous analysis, this would imply that
$k_1 = 0 \Rightarrow k_2 = 0 \Rightarrow k_3 = 0.$
From this analysis, it is immediate that
all three coefficients in
$[F_2]$ must be $0$.
Therefore:
(1) $~B = (-2/3)$ or $B = -1.$
(2) $~A = 0$ or $A = (-1/3).$
(3) $(3A+1)(B+3) + (3B+2)(A+2) = 0.$
$A = 0 \Rightarrow (B + 3) + 2(3B + 2) = 0
\Rightarrow 7B + 7 = 0 \Rightarrow B = -1.$
$A = (-1/3) \Rightarrow 0\times(B + 3) + (3B + 2) \times (5/3) = 0
\Rightarrow (3B + 2) = 0 \Rightarrow B = (-2/3).$
To satisfy the constraint that
$$ \frac{d}{du}\left[f(u) \times (g(u)\right] = \left[g(u)\right]^{(2/3)} $$
equation $[F_1]$ must be satisfied.
In $[F_1]$, setting $A = (-1/3), B = (-2/3)$ yields
$D \times 0 = 1$, which is impossible.
In $[F_1]$, setting $A = 0, B = -1$ yields
$D \times [S^2 - (-1)R^2] = 1 \Rightarrow $
[since $R^2 + S^2 = 2$] $D = (1/2).$
Consequently,
$$J ~=~ 3 \times \left[f(u) \times g(u)\right]^{(1/3)} $$
$$=~ 3 \times \left[RS^2 \times D^3 R^{(3A)} S^{(3B)}\right]^{(1/3)} $$
$$=~ 3 \times \left[RS^2 \times \left(\frac{1}{2}\right)^3 S^{(-3)}\right]^{(1/3)} $$
$$=~ \frac{3}{2} \times \left[RS^{(-1)}\right]^{(1/3)} ~
\text{which is symbolic for} $$
$$=~ \frac{3}{2} \times \left[(\sin u - \cos u) \times (\sin u + \cos u)^{(-1)}\right]^{(1/3)}. $$
$\underline{\text{Legitimacy of Approach}}$
Is this approach legitimate, or (instead) is it based on the known value of the closed form solution?
$\frac{d}{du} \left[f(u) \times g(u)\right]~$
will have two terms:
$$[f'(u) \times g(u)] ~\text{and} ~[f(u) \times g'(u)].$$
It seems that each term will have one or more factors $R$ and one or more
factors $S$. This suggests that the possibility of $g(u)$ containing one or more factors $R$ and/or one or more factors $S$ is worth exploring.
However, I've never seen any analysis that indicated that if an indefinite integral that resembled $J$ has a closed form solution, it must be limited to factors that resemble $R$ and $S$.
Therefore, a more universal approach is to employ functions $g_1(u), g_2(u),$ where:
$g_1(u) \times g_2(u) = g(u).$
$g_1(u)$ has form $D^3 R^{(3A)} S^{(3B)}$
$g_2(u)$ does not contain any factors $R$, or $S$, although it might contain factors $\sin (u)$ or $\cos (u)$.
In effect, my approach has simply hijacked the universal approach and blindly assumed that $g_2(u) = 1.$
It is arguable whether such a blind guess is a reasonable starting point for indefinite integrals whose form resembles $J$.
