Evaluate $\int \frac{dx}{(x-1)^{\frac 34} (x+2)^{\frac 54}}$

Question

Given that $\int \frac{1}{(x-1)^{\frac 34} (x+2)^{\frac 54}}dx$

Let $(x-1)^{\frac 14} = t$

So the integral becomes $$\int \frac{4}{(t^4+3)^{\frac 54}} dx$$

How do I solve it further?

Answer 1

Note that\begin{align}\frac1{(x-1)^{3/4}(x+2)^{5/4}}&=\frac{(x-1)^{-3/4}(x+2)^{-3/4}}{(x+2)^{2/4}}\\&=\frac{(x-1)^{-3/4}(x+2)^{-3/4}}{\bigl((x+2)^{1/4}\bigr)^2}.\end{align}On the other hand,$$(x-1)^{-3/4}=4\bigl((x-1)^{1/4}\bigr)'\quad\text{and}\quad(x+2)^{-3/4}=4\bigl((x+1)^{1/4}\bigr)',$$from which it follows that\begin{align}(x-1)^{-3/4}(x+2)^{-3/4}&=\frac13(x-1)^{-3/4}(x+2)^{-3/4}\bigl((x+2)-(x-1)\bigr)\\&=\frac13\bigl((x-1)^{-3/4}(x+2)^{1/4}-(x-1)^{1/4}(x+2)^{-3/4}\bigr)\\&=\frac43\left((x+2)^{1/4}\bigl((x-1)^{1/4}\bigr)'-(x-1)^{1/4}\bigl((x+1)^{1/4}\bigr)'\right).\end{align}Therefore,\begin{align}\frac1{(x-1)^{3/4}(x+2)^{5/4}}&=\frac{\frac43\left((x+2)^{1/4}\bigl((x-1)^{1/4}\bigr)'-(x-1)^{1/4}\bigl((x+1)^{1/4}\bigr)'\right)}{\bigl((x+2)^{1/4}\bigr)^2}\\&=\frac43\left(\frac{(x-1)^{1/4}}{(x+2)^{1/4}}\right)'.\end{align}So,$$\int\frac1{(x-1)^{3/4}(x+2)^{5/4}}\,\mathrm dx=\frac43\frac{(x-1)^{1/4}}{(x+2)^{1/4}}.$$

Answer 2

Continue with

$$I=\int \frac{4}{(t^4+3)^{\frac 54}} dt$$

and substitute $$\frac3{u^4}=1+\frac3{t^4}\>\>\> \implies \>\>\>t^4+3= \frac9{3-u^4},\>\>\>dt = \frac{3^{\frac54}}{(3-u^4)^{\frac54}}du $$

Then

$$I= \frac4{3^{\frac54}}\int du=\frac {4t}{3(t^4+3)^{\frac14}}+C $$