The integral $$\int \frac{1}{\sqrt[4]{(x-1)^3(x+2)^5}} dx$$ is equal to: (where $$C$$ is a constant of integration)

To make this integral manageable, we first rewrite the term inside the radical to create a common ratio:

$$(x-1)^3(x+2)^5 = (x+2)^8 \cdot \frac{(x-1)^3}{(x+2)^3} = (x+2)^8 \left( \frac{x-1}{x+2} \right)^3$$

Substitute this back into the integral:

$$I = \int \frac{1}{\sqrt[4]{(x+2)^8 \left( \frac{x-1}{x+2} \right)^3}} \, dx$$

$$I = \int \frac{1}{(x+2)^2 \left( \frac{x-1}{x+2} \right)^{3/4}} \, dx$$

Let $$t = \frac{x-1}{x+2}$$. To find $$dt$$, we use the quotient rule:

$$\frac{dt}{dx} = \frac{(x+2)(1) - (x-1)(1)}{(x+2)^2} = \frac{3}{(x+2)^2}$$

$$\frac{1}{3} \, dt = \frac{dx}{(x+2)^2}$$

Now, substitute $$t$$ and $$dt$$ into the integral:

$$I = \frac{1}{3} \int t^{-3/4} \, dt$$

Apply the power rule $$\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$$:

$$I = \frac{1}{3} \left( \frac{t^{1/4}}{1/4} \right) + C$$

$$I = \frac{4}{3} t^{1/4} + C$$

Back-substituting $$t = \frac{x-1}{x+2}$$ gives:

$$I = \frac{4}{3} \left( \frac{x-1}{x+2} \right)^{1/4} + C$$

So The Correct Option is B