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

The form of the integrand suggests the ansatz $$F(x) = c (x - a)^{1 + p} (x - b)^{1 + q}$$ for the antiderivative, where in our case $a = 1, b = -2$. Differentiating gives $$F'(x) = c (x - a)^p (x - b)^q [(2 + p + q) x - (a (q + 1) + b (p + 1))] .$$ When the linear term in brackets vanishes, we have $p + q = -2$, so we can write $p = -1 + r$, $q = -1 - r$ (as in our case, where $p = -\frac34, q = -\frac54$, so $r = \frac14$), giving the general formula (which applies when $a \neq b$ and $r \neq 0$) $$\boxed{\int (x - a)^{-1 + r} (x - b)^{-1 - r} \,dx = \frac{1}{r (a - b)} \left(\frac{x - a}{x - b}\right)^{r} + C} .$$

In our case, $$\int (x - 1)^{-\frac34} (x + 2)^{-\frac54} \,dx = \frac43 \left(\frac{x - 1}{x + 2}\right)^{\frac 14} + C.$$

If $p + q \neq -2$, we can express the antiderivative of $(x - a)^p (x - b)^q$ in terms of the ordinary hypergeometric function, ${}_2 F_1$.

Answer 2

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

Here's your generalized result specific to such kinds of integrals; $$I=\int \frac{1}{(x+a)^{\frac{m}{n}}(x+b)^{\frac{p}{n}}}\,dx$$

Where $\frac{m}{n}+\frac{p}{n}=2$

Factor out $(x+a)$,

$$I=\int \frac{1}{(x+a)^{2}\left(\frac{x+b}{x+a}\right)^{\frac{p}{n}}}\,dx$$

Just to visualize;

$$I=\int \frac{1}{(x+a)^{\frac{m}{n}+\frac{p}{n}}\left(\frac{x+b}{x+a}\right)^{\frac{p}{n}}}\,dx$$

Put $\left(\frac{x+b}{x+a}\right)=t$ and $\frac{1}{a-b}dt=\frac{1}{(x+a)^2}dx$

$$I(a;b;m;p;n) = \frac{1}{a-b} \int \frac{1}{t^{\frac{p}{n}}} dt$$

$$I(a;b;m;p;n) = \frac{1}{a-b} \left[\frac{t^{\frac{-p}{n}+1}}{\frac{-p}{n}+1}\right]+c$$

$$I(a;b;m;p;n) = \frac{1}{a-b} \left[\frac{t^{\frac{-p}{n}+1}}{\frac{-p}{n}+1}\right]+c$$

$$I(a;b;m;p;n)=\frac{n}{(n-p)(a-b)} \left[\frac{x+b}{x+a}\right]^{\frac{n-p}{n}}+c$$

The original question will then be; $$I(-1;2;3;5;4)=\frac43\left(\frac{x-1}{x+2}\right)^{\frac14}+C$$

Answer 3

Substitute $\frac3{u^4}=1+\frac3{t^4} $

$$I=\int \frac{4}{(t^4+3)^{\frac 54}} dt = \frac4{3^{\frac54}}\int du=\frac {4t}{3(t^4+3)^{\frac14}}+C $$

Answer 4

Using $x-1=t(x+2)$ seems to simplify the matters. With this substitution you obtain $$x=-\frac{2t+1}{t-1}$$ such that $$\int\frac{dx}{(x-1)^{3/4} (x+2)^{5/4}}=\int\frac{dt}{3 t^{3/4}}$$

Answer 5

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 6

Let $x=1+\dfrac1t\implies \mathrm dx=-\dfrac1{t^2}\,\mathrm dt$

\begin{align}I&=\int\dfrac1{(x-1)^{\frac34}(x+2)^{\frac54}}\,\mathrm dx\\&=\int\dfrac1{\left(\frac1t\right)^{\frac34}\left(3+\frac1t\right)^{\frac54}}\cdot\left(-\dfrac1{t^2}\right)\,\mathrm dt\\&=\int-t^{-\frac54}\left(3+\frac1t\right)^{-\frac54}\,\mathrm dt\\&=\int-(3t+1)^{-\frac54}\,\mathrm dt\\&=\dfrac43(3t+1)^{-\frac14}+C\\&=\dfrac43\left(\dfrac{3}{x-1}+1\right)^{-\frac14}+C\\&=\dfrac43\left(\dfrac{x+2}{x-1}\right)^{-\frac14}+C\\&=\boxed{\dfrac43\sqrt[4]{\dfrac{x-1}{x+2}}+C}\end{align}