Problem with integration by substitution - can't understand how $u$ was chosen.

Question

This is the problem and the solution to it:

$$\begin{split} \int \frac{1}{x-\sqrt{x}}dx& \\ u=\sqrt{x}-1&\quad du=\frac{1}{2\sqrt{x}}dx\\ \int \frac{1}{x-\sqrt{x}}dx &=2\int\frac1u du\\ &=2\ln|u|+C\\ &=2\ln|\sqrt{x}-1|+C\\ \end{split}$$

The problem is from Thomas' Calculus, 14th edition.

What I don't understand is, how was $u$ chosen and, more importantly, how was is replaced in the integral ? I understand how $dx$ was calculated. What I don't understand is how the author came up with $u = \sqrt{x}-1$ and how $u$ was then replaced inside the integral.

Answer 1

The choice that was made is not the only possible choice, although it is perhaps the more "efficient" one.

A more obvious choice would be to select $$u = \sqrt{x}, \quad du = \frac{1}{2\sqrt{x}} \, dx.$$ Equivalently, we may write this as $$x = u^2, \quad dx = 2u \, du.$$ This gives

$$\frac{1}{x - \sqrt{x}} \, dx = \frac{1}{u^2 - u} \cdot 2u\, du = \frac{2}{u - 1} \, du.$$ The remaining calculation of the integral is straightforward.

The motivation for such a choice is that we want to "remove" the square root on $x$, since it is generally easier to integrate expressions that are integer powers of the variable of integration. That said, the difference between $u = \sqrt{x}$ and $u = \sqrt{x} - 1$ is minor, since the latter substitution implies $$x = (u+1)^2, \quad dx = 2(u+1) \, du$$ and the only difference between the two is that there is an extra translation factor of $1$. The resulting integrand becomes $$\frac{1}{x - \sqrt{x}} \, dx = \frac{1}{(u+1)^2 - (u+1)} \cdot 2(u+1) \, du = \frac{2}{u} \, du.$$ In essence, the text's substitution can be thought of as two substitution steps combined into one, namely

$$u = \sqrt{x}, \quad v = u-1.$$

Indeed, if we were to integrate $2/(u-1)$ we would substitute $v = u-1$, $dv = du$, and write $$\int \frac{2}{u-1} \, du = \int \frac{2}{v} \, dv = 2 \log |v| + C = 2 \log |u-1| + C = 2 \log |\sqrt{x} - 1| + C.$$

Answer 2

Note how below serves as a motivation $$\frac1{x-\sqrt x} dx= \frac1{\sqrt x{(\sqrt x-1)}}dx= \frac{2\ d({\sqrt x)}}{{\sqrt x-1}}= \frac{2\ d({\sqrt x-1)}}{{\sqrt x-1}} $$

Answer 3

There are options for substitution such as $x=\tan \theta$. $$ \begin{aligned} I & =\int \frac{1}{\tan ^2 \theta-\tan \theta} \cdot 2 \tan \theta \sec ^2 \theta d \theta \\ & =2 \int \frac{1}{\tan \theta-1} d(\tan \theta) \\ & =2 \ln |\tan \theta-1|+C \\ & =2 \ln |\sqrt{x}-1|+C \end{aligned} $$ Therefore substitution are sometimes artistic.