The integral $$\int \frac{dx}{(x+1)^{3/4}(x-2)^{5/4}}$$, is equal to

We wish to evaluate $$\displaystyle\int \frac{dx}{(x-1)^{3/4}(x-2)^{5/4}}$$.

Rewrite the denominator by factoring out $$(x-2)^2$$: since $$(x-2)^2 \cdot \left(\dfrac{x-1}{x-2}\right)^{3/4} = (x-1)^{3/4}(x-2)^{5/4}$$, the integral becomes $$\displaystyle\int \frac{dx}{(x-2)^2 \cdot \left(\dfrac{x-1}{x-2}\right)^{3/4}}$$.

Substitute $$t = \dfrac{x-1}{x-2}$$. Then $$\dfrac{dt}{dx} = \dfrac{(x-2) - (x-1)}{(x-2)^2} = \dfrac{-1}{(x-2)^2}$$, so $$\dfrac{dx}{(x-2)^2} = -dt$$.

The integral transforms to $$\displaystyle\int \frac{-dt}{t^{3/4}} = -\int t^{-3/4}\,dt = -\frac{t^{1/4}}{1/4} + c = -4\,t^{1/4} + c = -4\left(\frac{x-1}{x-2}\right)^{\!1/4} + c$$.