Evaluate the integral: I = integral dx/(6th root

Evaluate the integral: $$ I = \int \frac{dx}{\sqrt[6]{(x-3)^2(x+1)^5}} = \int (x-3)^{-1/3}(x+1)^{-5/6}\,dx. $$

Step 1 · Substitute

Let $t=\dfrac{x-3}{x+1}$, so $x=\dfrac{3+t}{1-t}$, $dx=\dfrac4{(1-t)^2}dt$, $x-3=\dfrac{4t}{1-t}$, $x+1=\dfrac4{1-t}$. Substituting:

$$ I = 4^{-1/6}\int t^{-1/3}(1-t)^{-5/6}\,dt. $$

Step 2 · Recognize the exponent obstruction

Here $p=-\tfrac13$, $q=-\tfrac56$, $p+q=-\tfrac76$ — none is an integer, so no rational substitution can reduce this to a rational-function integral. This is the same obstruction as in Problem 329.

Step 3 · Hypergeometric closed form

Using $\displaystyle\int t^{\alpha-1}(1-t)^{\beta-1}dt=\frac{t^\alpha}\alpha\,{}_2F_1(\alpha,1-\beta;\alpha+1;t)$ with $\alpha=\tfrac23$ (so $\alpha-1=-\tfrac13$) and $\beta=\tfrac16$ (so $\beta-1=-\tfrac56$), giving $1-\beta=\tfrac56$, $\alpha+1=\tfrac53$:

$$ \int t^{-1/3}(1-t)^{-5/6}\,dt = \frac32\,t^{2/3}\,{}_2F_1\left(\frac23,\frac56;\frac53;t\right) + C. $$

Summary Table

Additional Notes

• Major rewrite — genuinely non-elementary: the source's Step 8 abandoned rigor ("let's check if the derivative... this doesn't match... after careful evaluation") and asserted the unjustified boxed answer $3\sqrt[4]{(x-3)/(x+1)}$, which fails numerically (derivative $\approx0.401$ vs. true integrand $\approx0.262$ at $x=4$). The exponent analysis above shows no elementary closed form exists; the hypergeometric expression is the best available, and it was verified to match the integrand exactly.
• The domain is $x>-1$ (needed for $(x+1)^5\geq0$ under the sixth root); the branch $t=(x-3)/(x+1)\in(0,1)$ corresponds to $x>3$.
• The result was verified numerically at $x=4,5,10$, with the derivative of the boxed formula matching the integrand $(x-3)^{-1/3}(x+1)^{-5/6}$ exactly at all three points (e.g. $\approx0.2615$ at $x=4$).