Example 36: $\int\frac{(x^4-x)^{1/4}}{x^5}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   We need $x^4 - x \ge 0$ for the fourth root and $x \ne 0$ for the division by $x^5$⁠. Factor:

   $$x^4 - x = x\,(x^3 - 1) .$$

   The product is $\ge 0$ when both factors have the same sign or one is 0: $x \ge 1$ (both $\ge 0$⁠) or $x \le 0$ (both $\le 0$⁠). Removing $x = 0$⁠, the domain is

   $$x < 0 \quad\text{or}\quad x \ge 1 .$$
   An indefinite integral is an antiderivative on an interval where the integrand is defined. Here there are two separate intervals, $(-\infty, 0)$ and $[1, \infty)$⁠, and we must treat each one. Between them, for $0 < x < 1$⁠, $x^4 - x < 0$ has no real fourth root.
   $x^3 - 1 \ge 0 \iff x \ge 1$⁠, because $x^3$ is increasing.

2. For $x \ge 1$⁠: take $x^4$ out of the root

   For $x > 0$⁠, $x^4 - x = x^4\left(1 - x^{-3}\right)$ and $(x^4)^{1/4} = x$⁠, so

   $$(x^4 - x)^{1/4} = x\left(1 - \frac{1}{x^3}\right)^{1/4} .$$

   Dividing by $x^5$⁠:

   $$\begin{aligned}\frac{(x^4 - x)^{1/4}}{x^5} &= \frac{x\,(1 - x^{-3})^{1/4}}{x^5}\\ &= \frac{(1 - x^{-3})^{1/4}}{x^4} .\end{aligned}$$
   This is the book's first line. It turns the awkward root of a difference into a root of something whose derivative we can see. It uses $(x^4)^{1/4} = x$⁠, true only for $x > 0$⁠; Step 6 handles $x < 0$⁠. For $x \ge 1$⁠, $1 - x^{-3} \ge 0$⁠, so its fourth root is defined.
   $(pq)^{1/4} = p^{1/4}q^{1/4}$ for $p, q \ge 0$⁠.

3. Substitute $t = 1 - x^{-3}$

   Put $t = 1 - \frac{1}{x^3} = 1 - x^{-3}$⁠. Then

   $$\frac{dt}{dx} = -(-3)x^{-4} = \frac{3}{x^4}, \qquad\text{so}\qquad \frac{dx}{x^4} = \frac{1}{3}\,dt .$$

   The integral becomes

   $$\int \frac{(1 - x^{-3})^{1/4}}{x^4}\,dx = \frac{1}{3}\int t^{1/4}\,dt .$$
   The factor $\frac{1}{x^4}$ left over from Step 2 is exactly what $dt$ needs, apart from the constant 3. So after the substitution no $x$ remains. For $x \ge 1$⁠, $t$ runs over $[0, 1)$⁠.
   $\frac{d}{dx}x^{-3} = -3x^{-4}$⁠, so $\frac{d}{dx}(1 - x^{-3}) = 3x^{-4}$⁠. Lesson: integration by substitution

4. Integrate and substitute back

   $$\begin{aligned}\frac{1}{3}\int t^{1/4}\,dt &= \frac13\cdot\frac{t^{5/4}}{5/4} + C\\ &= \frac13\cdot\frac45\,t^{5/4} + C = \frac{4}{15}\,t^{5/4} + C .\end{aligned}$$

   With $t = 1 - \frac{1}{x^3}$⁠, for $x \ge 1$⁠:

   $$\int \frac{(x^4-x)^{1/4}}{x^5}\,dx = \frac{4}{15}\left(1 - \frac{1}{x^3}\right)^{5/4} + C .$$
   Power rule with $n = \frac14$⁠: add 1 to get $\frac54$⁠, divide by $\frac54$ (multiply by $\frac45$⁠). This is the textbook's answer, and it is correct on $x \ge 1$ (at $x = 1$ the derivative is one-sided, from the right).
   Standard formula (i), Section 7.2.

5. Check: $\frac{d}{dx}$ of the answer (for $x > 1$⁠)

   Chain rule with outer $u^{5/4}$ and inner $u = 1 - x^{-3}$⁠, $u' = 3x^{-4}$⁠:

   $$\begin{aligned}\frac{d}{dx}\left[\frac{4}{15}u^{5/4}\right] &= \frac{4}{15}\cdot\frac54\,u^{1/4}\cdot\frac{3}{x^4}\\ &= \frac{(1 - x^{-3})^{1/4}}{x^4} .\end{aligned}$$

   Multiplying top and bottom by $x$ (with $x\,(1 - x^{-3})^{1/4} = (x^4 - x)^{1/4}$ for $x > 0$⁠, Step 2) gives $\frac{(x^4 - x)^{1/4}}{x^5}$⁠, the integrand.

   $\frac{4}{15}\cdot\frac54\cdot 3 = 1$⁠: the constants cancel exactly, as they must.

6. For $x < 0$⁠: the sign changes

   For $x < 0$⁠, $(x^4)^{1/4} = |x| = -x$⁠, so

   $$(x^4 - x)^{1/4} = -x\left(1 - x^{-3}\right)^{1/4} ,$$

   (here $1 - x^{-3} > 1$⁠, so the root is fine) and the integrand is

   $$\frac{(x^4-x)^{1/4}}{x^5} = -\frac{(1 - x^{-3})^{1/4}}{x^4} .$$

   The same substitution then gives, for $x < 0$⁠,

   $$\int \frac{(x^4-x)^{1/4}}{x^5}\,dx = -\frac{4}{15}\left(1 - \frac{1}{x^3}\right)^{5/4} + C .$$
   For negative $x$ the integrand is negative (positive root over the negative $x^5$⁠), but the book's answer $\frac{4}{15}(1 - x^{-3})^{5/4}$ has the positive derivative $\frac{(1 - x^{-3})^{1/4}}{x^4}$⁠. So the book's formula is right only on $x \ge 1$⁠; on $x < 0$ it needs a minus sign. For example at $x = -1$ the integrand is $\frac{2^{1/4}}{-1} \approx -1.189207$⁠, while the derivative of the book's formula there is $+1.189207$⁠.
   $|x| = -x$ when $x < 0$⁠. Basics: powers and roots

7. One formula for the whole domain

   On the domain ($x \ne 0$⁠),

   $$\begin{aligned}\left(1 - \frac{1}{x^3}\right)^{5/4} &= \frac{(x^4 - x)^{5/4}}{(x^4)^{5/4}}\\ &= \frac{(x^4-x)^{5/4}}{|x|^5} .\end{aligned}$$

   Multiplying by the sign of $x$ turns $|x|^5$ into $x^5$⁠, so Steps 4 and 6 combine into

   $$\int \frac{(x^4-x)^{1/4}}{x^5}\,dx = \frac{4}{15}\cdot\frac{(x^4 - x)^{5/4}}{x^5} + C ,$$

   valid on both pieces of the domain, $x < 0$ and $x \ge 1$ (with an independent constant on each piece).

   For $x \ge 1$⁠, $x^5 = |x|^5$ and this is the book's form; for $x < 0$⁠, $x^5 = -|x|^5$ and this is Step 6. Writing it this way keeps the sign automatically correct.

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