Exercise 7.2 · Q15: $\int\frac{x}{9-4x^2}\,dx$

Basics you need

Solution

1. Compare the numerator with the derivative of the denominator

   $$\frac{d}{dx}\left(9 - 4x^2\right) = -8x.$$

   The numerator is $x$⁠, which is $-\frac18$ times $-8x$⁠. So, apart from a constant factor, the numerator is the derivative of the denominator.

   Constant factors never matter in a substitution, because they can be moved outside the integral. What matters is that the variable part, $x$⁠, matches. If the numerator were $1$ instead of $x$⁠, this substitution would fail; that is a different integral, handled by a formula of Section 7.4.
   $\frac{d}{dx}(9) = 0$ and $\frac{d}{dx}(4x^2) = 8x$ (power rule). Basics: the power rule

2. Substitute $t = 9 - 4x^2$

   Let $t = 9 - 4x^2$⁠. Then $dt = -8x\,dx$⁠, so

   $$x\,dx = -\frac{1}{8}\,dt.$$

   Therefore

   $$\begin{aligned} \int \frac{x}{9 - 4x^2}\,dx &= \int \frac{1}{t}\left(-\frac18\,dt\right) \\ &= -\frac18\int\frac{dt}{t}. \end{aligned}$$
   We solved $dt = -8x\,dx$ for the exact product $x\,dx$ that appears in the integrand, then swapped it in. The constant $-\frac18$ comes out in front by Property (IV). The new integral has no $x$ left in it, so the substitution is complete.
   Change of variable, Section 7.3.1. Lesson: integration by substitution

3. Integrate $\frac1t$

   $$-\frac18\int\frac{dt}{t} = -\frac18\log|t| + C.$$
   The modulus is essential here: $t = 9 - 4x^2$ is negative whenever $|x| > \frac32$ (for example $x = 2$ gives $t = -7$⁠), and the logarithm of a negative number is not defined. The formula $\log|t|$ works for every $t \ne 0$⁠.
   Standard formula (xii), Section 7.2: $\int\frac1x\,dx = \log|x| + C$⁠. Basics: logarithms and 1/x

4. Return to $x$

   $$\int \frac{x}{9 - 4x^2}\,dx = -\frac18\log\left|9 - 4x^2\right| + C.$$
   Replace $t$ by $9 - 4x^2$ so that the answer is in the original variable. The modulus stays, because $9 - 4x^2$ is positive for $|x| < \frac32$ and negative for $|x| > \frac32$⁠.

5. Check: differentiate the answer back

   For $u \ne 0$⁠, $\frac{d}{dx}\log|u| = \frac{u'}{u}$ (formula (xii) of Section 7.2 together with the chain rule). With $u = 9 - 4x^2$⁠, $u' = -8x$⁠:

   $$\begin{aligned} \frac{d}{dx}\left[-\frac18\log\left|9 - 4x^2\right|\right] &= -\frac18\cdot\frac{-8x}{9 - 4x^2} \\ &= \frac{x}{9 - 4x^2}. \end{aligned}$$

   This is the integrand, so the answer is correct.

   The two minus signs and the two eights cancel. Checking by differentiation catches the most common slip in this question, a missing minus sign.
   Property (I), Section 7.2.1: $\frac{d}{dx}\int f(x)\,dx = f(x)$⁠. Basics: integration undoes differentiation

6. Where the answer is valid

   The integrand and the answer are both defined exactly when $9 - 4x^2 \ne 0$⁠, that is, $x \ne -\frac32$ and $x \ne \frac32$⁠. These two points cut the real line into three intervals:

   $$x < -\tfrac32, \qquad -\tfrac32 < x < \tfrac32, \qquad x > \tfrac32.$$

   The answer holds on each of them.

   An antiderivative is found on an interval. Strictly, the constant $C$ may take a different value on each of the three intervals, because a function with zero derivative is constant only on a single interval. The book does not usually mention this (Note, Section 7.2), but it is worth knowing.

Every fact used, and where it comes from