Miscellaneous · Q4: $\int\frac{dx}{x^2(x^4+1)^{3/4}}$

Basics you need

Solution

1. Why this method

   The integrand is $x^{-2}(x^4+1)^{-3/4}$⁠. Trying $t = x^4 + 1$ would need a factor $x^3$ in the numerator, which is not there. But if we take $x^4$ out of the bracket, the bracket becomes $1 + x^{-4}$⁠, whose derivative $-4x^{-5}$ is a power of $x$⁠, and the powers of $x$ outside combine to exactly $x^{-5}$⁠.

   Count the powers: outside we have $x^{-2}$⁠, and $(x^4)^{-3/4}$ contributes $x^{-3}$ (for $x > 0$⁠), giving $x^{-5}$ in total. Since $\frac{d}{dx}(1 + x^{-4}) = -4x^{-5}$⁠, the substitution matches perfectly. In general, this "take the highest power out" trick works when the powers of $x$ left outside the bracket combine to $x^{-(n+1)}$⁠, which is (up to a constant factor) the derivative of the $x^{-n}$ inside the new bracket.
   Substitution, Section 7.3.1. Lesson: integration by substitution

2. For $x > 0$⁠: take $x^4$ out of the bracket

   For $x > 0$⁠, $(x^4)^{3/4} = x^3$⁠, so

   $$\begin{aligned} (x^4+1)^{3/4} &= \Big[x^4\Big(1 + \frac{1}{x^4}\Big)\Big]^{3/4} \\ &= x^3\Big(1 + \frac{1}{x^4}\Big)^{3/4}. \end{aligned}$$

   Hence

   $$f(x) = \frac{1}{x^5\,(1 + x^{-4})^{3/4}}.$$
   Both $x^4$ and $1 + x^{-4}$ are positive, so the power of the product splits into the product of powers. The fourth root of $x^4$ is $|x|$⁠, which equals $x$ only because $x > 0$ here; that is why we treat $x < 0$ separately in Step 4.
   Laws of indices (Basic 1).

3. Substitute $t = 1 + x^{-4}$ and integrate

   Then $dt = -4x^{-5}\,dx$⁠, so $\dfrac{dx}{x^5} = -\dfrac{dt}{4}$⁠. Note $t > 1$⁠.

   $$\begin{aligned} \int f(x)\,dx &= \int (1 + x^{-4})^{-3/4}\,\frac{dx}{x^5} \\ &= -\frac14\int t^{-3/4}\,dt = -\frac14\cdot 4t^{1/4} + C \\ &= -\Big(1 + \frac{1}{x^4}\Big)^{1/4} + C. \end{aligned}$$
   After substituting, only a power of $t$ is left; the power rule with $n = -\frac34$ gives $\frac{t^{1/4}}{1/4}$⁠. The factor $-\frac14$ from $dt$ cancels the 4. This is the textbook's answer, and it is correct for $x > 0$⁠.
   Standard formula (i), Section 7.2; substitution, Section 7.3.1.

4. The case $x < 0$

   Now $(x^4)^{3/4} = |x|^3 = -x^3$⁠, so Step 2 becomes $(x^4+1)^{3/4} = -x^3(1 + x^{-4})^{3/4}$ and

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

   The same substitution now gives the opposite sign:

   $$\int f(x)\,dx = \Big(1 + \frac{1}{x^4}\Big)^{1/4} + C \quad (x < 0).$$
   The integrand is positive for every $x \ne 0$⁠, so any antiderivative must be increasing on each interval. On $x < 0$⁠, as $x$ increases towards 0, $|x|$ shrinks, so $1 + \frac{1}{x^4}$ grows (for example $x = -2$ gives $1.0625$ and $x = -1$ gives 2). So $+(1 + x^{-4})^{1/4}$ increases there, as it must, while the key's $-(1 + x^{-4})^{1/4}$ decreases. The key's form therefore holds only for $x > 0$⁠.
   $|x| = -x$ for $x < 0$ (Basic 1).

5. One formula for both signs, and the key's form

   For any $x \ne 0$⁠, $\big(1 + \frac{1}{x^4}\big)^{1/4} = \dfrac{(x^4+1)^{1/4}}{|x|}$⁠. So Steps 3 and 4 combine into

   $$\int\frac{dx}{x^2(x^4+1)^{3/4}} = -\frac{(x^4+1)^{1/4}}{x} + C.$$

   For $x > 0$ this is $-\frac{(x^4+1)^{1/4}}{|x|} = -\big(1 + \frac{1}{x^4}\big)^{1/4}$⁠, identical to the key's form.

   Dividing by $x$ instead of $|x|$ supplies the minus sign automatically when $x < 0$⁠. On each of the two intervals $x > 0$ and $x < 0$ the constant $C$ can be different.
   $\big(\frac{p}{q}\big)^{1/4} = \frac{p^{1/4}}{q^{1/4}}$ for $p, q > 0$⁠, with $(x^4)^{1/4} = |x|$⁠.

6. Check: differentiate the answer

   Quotient rule on $F = -\dfrac{(x^4+1)^{1/4}}{x}$⁠, using $\frac{d}{dx}(x^4+1)^{1/4} = x^3(x^4+1)^{-3/4}$⁠:

   $$\begin{aligned} F' &= -\frac{x\cdot x^3(x^4+1)^{-3/4} - (x^4+1)^{1/4}}{x^2} \\ &= -\frac{(x^4+1)^{-3/4}\big[x^4 - (x^4+1)\big]}{x^2} \\ &= \frac{1}{x^2(x^4+1)^{3/4}}. \ \checkmark \end{aligned}$$
   In the second line we took out the common factor $(x^4+1)^{-3/4}$⁠, using $(x^4+1)^{1/4} = (x^4+1)^{-3/4}(x^4+1)$⁠. The bracket is $-1$⁠, which cancels the minus sign in front. This works for every $x \ne 0$⁠, with no sign assumption.
   Quotient rule and chain rule, Chapter 5. Basics: integration undoes differentiation

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