Miscellaneous · Q5: $\int\frac{dx}{x^{1/2}+x^{1/3}}$

Basics you need

Solution

1. Why the substitution $x = t^6$

   We want a power of $t$ whose square root and cube root are both whole powers. That needs an exponent divisible by both 2 and 3, and the smallest is 6. On $x > 0$ put $x = t^6$ with $t = x^{1/6} > 0$⁠. This is one-one and differentiable, and covers all $t > 0$⁠.

   With roots of different orders, no single standard formula applies, and the substitution $t = \sqrt x$ would leave $x^{1/3} = t^{2/3}$ behind. The hint's factorisation $x^{1/2} + x^{1/3} = x^{1/3}(1 + x^{1/6})$ shows that $x^{1/6}$ is the natural new variable, which is the same as $x = t^6$⁠.
   Least common multiple of 2 and 3 is 6; fractional powers (Basic 1).

2. Substitute

   $dx = 6t^5\,dt$⁠, $x^{1/2} = t^3$⁠, $x^{1/3} = t^2$⁠. So

   $$\begin{aligned} \int\frac{dx}{x^{1/2}+x^{1/3}} &= \int\frac{6t^5\,dt}{t^3 + t^2} \\ &= \int\frac{6t^5}{t^2(t+1)}\,dt = 6\int\frac{t^3}{t+1}\,dt. \end{aligned}$$
   Cancelling $t^2$ is allowed because $t > 0$⁠. The result is a rational function of $t$⁠, and an improper one: the top has degree 3, the bottom degree 1.
   Substitution, Section 7.3.1. Lesson: integration by substitution

3. Divide $t^3$ by $t + 1$

   Since $t^3 + 1 = (t+1)(t^2 - t + 1)$ (sum of cubes), $t^3 = (t+1)(t^2 - t + 1) - 1$⁠. Dividing by $t + 1$⁠:

   $$\frac{t^3}{t+1} = t^2 - t + 1 - \frac{1}{t+1}.$$
   Long division of $t^3$ by $t + 1$ gives the same quotient $t^2 - t + 1$ and remainder $-1$⁠; the sum-of-cubes identity is a quick way to see it. Check: $(t+1)(t^2 - t + 1) = t^3 + 1$⁠, and $t^3 + 1 - 1 = t^3$⁠. ✓
   Division of polynomials, Section 7.5. Basics: splitting fractions

4. Integrate in $t$

   $$\begin{aligned} 6\int\frac{t^3}{t+1}\,dt &= 6\Big[\frac{t^3}{3} - \frac{t^2}{2} + t - \log(1+t)\Big] + C \\ &= 2t^3 - 3t^2 + 6t - 6\log(1+t) + C. \end{aligned}$$
   Each power of $t$ uses the power rule, and $\int\frac{dt}{t+1} = \log|t+1|$ (standard formula (xii) with $t + 1$ in place of $t$⁠). Since $t > 0$⁠, $t + 1 > 0$ and the absolute value can be dropped.
   Standard formulae (i) and (xii), Section 7.2; properties (III), (IV), Section 7.2.1. Basics: logarithms and 1/x

5. Return to $x$

   Put $t = x^{1/6}$⁠: $t^3 = x^{1/2} = \sqrt x$⁠, $t^2 = x^{1/3}$⁠. So

   $$\begin{aligned} \int\frac{dx}{x^{1/2}+x^{1/3}} &= 2\sqrt x - 3x^{1/3} + 6x^{1/6} \\ &\quad - 6\log\big(1 + x^{1/6}\big) + C. \end{aligned}$$
   This is exactly the textbook's answer, term for term.
   Fractional powers (Basic 1).

6. Check: differentiate the answer

   Differentiate in $t$ first, then use $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{6t^5}$ (chain rule). With $F = 2t^3 - 3t^2 + 6t - 6\log(1+t)$⁠:

   $$\begin{aligned} \frac{dF}{dt} &= 6t^2 - 6t + 6 - \frac{6}{1+t} \\ &= \frac{6\big[(t^2 - t + 1)(1+t) - 1\big]}{1+t} = \frac{6t^3}{1+t}. \end{aligned}$$

   So

   $$\frac{dF}{dx} = \frac{6t^3}{1+t}\cdot\frac{1}{6t^5} = \frac{1}{t^2(1+t)} = \frac{1}{t^3 + t^2}.$$

   With $t^3 = x^{1/2}$ and $t^2 = x^{1/3}$ this is $\dfrac{1}{x^{1/2}+x^{1/3}}$⁠. ✓

   The bracket uses $(t^2 - t + 1)(1+t) = t^3 + 1$ again. Going through $t$ is legitimate because $t = x^{1/6}$ is differentiable for $x > 0$ with $\frac{dx}{dt} = 6t^5 \ne 0$⁠.
   Chain rule, Chapter 5. Basics: integration undoes differentiation

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