Integrals in the form $\int R(x,x^{\frac{m}{n}},x^{\frac{r}{s}},...)$

Question

How to approach integrals in the form $\int R(x,x^{\frac{m}{n}},x^{\frac{r}{s}}...)dx$

where $R(x,x^{\frac{m}{n}},x^{\frac{r}{s}})$ represents a fraction in terms of roots:

I was told to use the substitution $x=t^k$ where k is the least common multiple of the denominators of exponents of x

let's do an example:

$$\int \frac{dx}{x-\sqrt{x}}$$

$lcm(1,2) = 2$

hence $x=t^2$ and $dx=2tdt$ then:

$$\int\frac{2tdt}{t^2-t}$$ $$2\int\frac{dt}{t-1} = 2ln|(t-1)|=2ln|\sqrt{x}-1|$$

I was told in my class to use this method but I was never told WHY it works I would like to have a deeper understanding of why this method works and also see if there are more methods to approach integrals of that form

Answer 1

If you do this substitution, then you always end up with a rational function, because of how you chose the power $k$. But thanks to the partial fraction decomposition we have an algorithm to integrate any rational function (provided that we can factorize its denominator). See this MSE post for more details.

In answer to the comment: $R$ takes as arguments $x^{\frac mn}$, $x^{\frac rs}$, and some other rational powers of $x$. Since $k$ is the least common multiple of $n,s$, and so on (it may as well be any common multiple of them), then after substituting $x = t^{k}$ we get $x^{\frac mn} = t^{m\cdot \frac kn}$, $x^{\frac rs} = t^{r\cdot \frac ks}$. Since $k$ is a multiple of $n$ and $s$ we have $m\cdot \frac kn, r\cdot \frac ks \in \mathbb{N}$, i.e., $t$ occurs only in natural powers. That's why after substitution you get a rational function.

Answer 2

Suppose we have a rational function $R(u_1, \ldots, u_m)$ and rational numbers $\frac{p_1}{q_1}, \ldots, \frac{p_m}{q_m}$, and denote $q := \operatorname{lcm}(q_1, \ldots, q_m)$. Then, we can rewrite each rational number $\frac{p_i}{q_i}$ as $\frac{r_i}{q}$ for some integer $r_i$. In terms of $t := x^q$, we have $x^\frac{r_i}{q} = (t^q)^{\frac{r_i}{q}} = t^{r_i}$, and $dx = q t^{q - 1} \,dt$, so that substitution gives $$\int R\left(x^\frac{p_1}{q_1}, \ldots, x^\frac{p_m}{q_m}\right) \,dx = q \int R(t^{r_1}, \ldots, t^{r_m}) t^{q - 1} \,dt .$$ In particular, the integrand of the integral in $t$ is a product of rational functions of $t$ and hence is itself rational in $t$. But (provided you can express the roots of its denominator) you can always evaluate an indefinite integral of a rational function in a closed form in terms of elementary functions.

Example We can generalize the example in the question statement: For any integer $q > 1$, the substitution $x = t^q$ gives $$\int \frac{dx}{x - x^{1 / q}} = q \int \frac{t^{q - 2} \,dt}{t^{q - 1} - 1},$$ and the substitution $u = t^{q - 1} - 1$ transforms the integral to $$\frac{q}{q - 1} \int \frac{du}{u} = \frac{q}{q - 1} \log |u| + C = \frac{q}{q - 1} \log \left\vert x^{(q - 1) / q} - 1\right\vert + C$$