Exercise 7.4 · Q5: $\int\frac{3x}{1+2x^4}\,dx$

Basics you need

Solution

1. Where is the integrand defined?

   $2x^4 \ge 0$⁠, so $1 + 2x^4 \ge 1$⁠: the denominator never vanishes. The integrand is defined for every real $x$⁠.

   No restriction is needed, so the answer will hold on the whole real line.

2. Substitute $t = x^2$

   Put $t = x^2$⁠. Then $dt = 2x\,dx$⁠, so $3x\,dx = \frac32\,dt$⁠, and $2x^4 = 2(x^2)^2 = 2t^2$⁠:

   $$\int\frac{3x\,dx}{1 + 2x^4} = \frac32\int\frac{dt}{1 + 2t^2}.$$
   The odd power $x$ in the numerator is the sign to try $t = x^2$⁠: it supplies the $x\,dx$ needed for $dt$⁠, and the $x^4$ becomes $t^2$⁠. The 3 and the $\frac12$ combine to $\frac32$⁠.
   Integration by substitution, Section 7.3.1. Lesson: integration by substitution

3. Take the 2 out of the denominator

   $$\begin{aligned} 1 + 2t^2 &= 2\left(t^2 + \frac12\right) \\ &= 2\left(t^2 + \left(\tfrac{1}{\sqrt2}\right)^2\right). \end{aligned}$$

   So

   $$\frac32\int\frac{dt}{1 + 2t^2} = \frac34\int\frac{dt}{t^2 + \left(\frac{1}{\sqrt2}\right)^2}.$$
   Formula 7.4 (3) needs $t^2$ with coefficient 1, so the 2 is factored out (as in type (7) of Section 7.4). The constant $\frac12$ left over is written as a square, $\left(\frac{1}{\sqrt2}\right)^2$⁠, to read off $a$⁠. Then $\frac32\cdot\frac12 = \frac34$⁠.

4. Use formula 7.4 (3) with $a = \frac{1}{\sqrt2}$

   Here $\frac1a = \sqrt2$ and $\frac ta = \sqrt2\,t$⁠, so

   $$\frac34\int\frac{dt}{t^2 + \left(\frac{1}{\sqrt2}\right)^2} = \frac34\cdot\sqrt2\,\tan^{-1}(\sqrt2\,t) + C.$$
   Dividing by $a = \frac{1}{\sqrt2}$ is the same as multiplying by $\sqrt2$⁠, both in front and inside $\tan^{-1}$⁠.
   Formula 7.4 (3). Lesson: integrals of particular functions

5. Return to $x$ and tidy the constant

   With $t = x^2$⁠, and $\frac{3\sqrt2}{4} = \frac{3\sqrt2}{2\cdot\sqrt2\cdot\sqrt2} = \frac{3}{2\sqrt2}$⁠,

   $$\int\frac{3x}{1 + 2x^4}\,dx = \frac{3}{2\sqrt2}\tan^{-1}(\sqrt2\,x^2) + C.$$
   $4 = 2\cdot 2 = 2\cdot\sqrt2\cdot\sqrt2$⁠, so one $\sqrt2$ cancels. Both $\frac{3\sqrt2}{4}$ and $\frac{3}{2\sqrt2}$ equal $1.0607$ (to 4 places); the second is the textbook's form. A shortcut gives the same in one step: $u = \sqrt2\,x^2$ has $du = 2\sqrt2\,x\,dx$ and $1 + 2x^4 = 1 + u^2$⁠, so the integral is $\frac{3}{2\sqrt2}\int\frac{du}{1 + u^2}$⁠.

6. Check: differentiate the answer back

   Let $F(x) = \frac{3}{2\sqrt2}\tan^{-1}u$ with $u = \sqrt2\,x^2$⁠, $u' = 2\sqrt2\,x$⁠. Then

   $$\begin{aligned} F'(x) &= \frac{3}{2\sqrt2}\cdot\frac{1}{1 + 2x^4}\cdot 2\sqrt2\,x \\ &= \frac{3x}{1 + 2x^4}. \ \checkmark \end{aligned}$$
   Chain rule with outer $\tan^{-1}u$ and inner $u = \sqrt2\,x^2$⁠; $u^2 = 2x^4$⁠. The factors $2\sqrt2$ cancel exactly.
   Standard formula (x), Section 7.2. Basics: integration undoes differentiation

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