1. ## Integration

$\int \frac{dx}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}$

2. Hint

\displaystyle{\begin{aligned}\frac{1}{(x-1)^{3/4}(x+2)^{5/4}}&=\frac{1}{(x-1)^{3/4}(x+2)^{2-3/4}}=\frac{1}{(x-1)^{3/4}\dfrac{(x+2)^2}{(x+2)^{3/4}}}=\\[2pt]&=\frac{1}{{\left(\dfrac{x-1}{x+2}\right)\!}^{3/4}(x+2)^2}={\left(\frac{x-1}{x+2}\right)\!}^{-3/4}}\frac{1}{(x+2)^2}\end{aligned}}

Now try to use this substitution $\dfrac{x-1}{x+2}=u~\Rightarrow~\dfrac{dx}{(x+2)^2}=\dfrac{d u}{3}$

3. Let $t = \frac{x-1}{x+2} \Rightarrow \frac{dt}{dx} = \frac{3}{(x+2)^2} \Rightarrow dx = \frac{1}{3}(x+2)^2\;{dt}$. Then we have:

\displaystyle \begin{aligned} \int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}\;{dx} & = \frac{1}{3}\int\frac{(x+2)^2}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}\;{dt} = \frac{1}{3}\int\frac{(x+2)^{\frac{3}{4}}}{(x-1)^{\frac{3}{4}}}\;{dt} \\& = \frac{1}{3}\int\left(\frac{x+2}{x-1}\right)^{\frac{3}{4}}\;{dt} = \frac{1}{3}\int\frac{1}{t^{\frac{3}{4}}}\;{dt} = \frac{4}{3}t^{\frac{1}{4}}+k. \end{aligned}

Thus $\displaystyle \int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{3}}}\;{dx} = \frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+k.$

EDIT: Beaten by four minutes. And I thought this solution was clever/unique!

4. Typos

$\displaystyle \int \frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{3}}}\;{dx} \ne \frac{1}{3}\int\frac{(x+2)^2}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{3}}}\;{dt} =\ldots$

5. Originally Posted by DeMath
Typos
4 instead of 3, right? If you're referring to that, then sorted, thanks.

6. Sorry, at first, I did not understand the course of your solution.

7. By parts, anyone? Do the addition/subtraction (how about 'numerator give-and-take'?!) thing -

\displaystyle{\begin{aligned}\frac{1}{(x - 1)^{3/4}(x+2)^{5/4}} & = \frac{x + 2 - x - 1}{(x - 1)^{3/4}(x+2)^{5/4}} \\& = \frac{x + 2}{(x - 1)^{3/4}(x+2)^{5/4}} - \frac{x + 1}{(x - 1)^{3/4}(x+2)^{5/4}} \\& = \frac{1}{(x - 1)^{3/4}(x+2)^{1/4}} - \frac{x + 1}{(x - 1)^{3/4}(x+2)^{5/4}}\end{aligned}}

Then do integration by parts on the first half...

... where (key in spoiler) ...

Spoiler:

... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,

... is lazy integration by parts, doing without u and v.

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8. Thank you very much to all..