Exercise 7.9 · Q6: $\int_0^2\frac{dx}{x+4-x^2}$

Basics you need

Solution

1. Complete the square and check the denominator

   $$\begin{aligned}x + 4 - x^2 &= 4 - \left(x^2 - x\right) \\ &= 4 - \left[\left(x - \tfrac{1}{2}\right)^2 - \tfrac{1}{4}\right] \\ &= \tfrac{17}{4} - \left(x - \tfrac{1}{2}\right)^2.\end{aligned}$$

   For $0 \le x \le 2$⁠, $-\frac{1}{2} \le x - \frac{1}{2} \le \frac{3}{2}$⁠, so $\left(x - \frac{1}{2}\right)^2 \le \frac{9}{4}$ and

   $$x + 4 - x^2 \ge \frac{17}{4} - \frac{9}{4} = 2 > 0.$$

   So the integrand is continuous on $[0, 2]$⁠.

   The denominator is 0 at $x = \frac{1 \pm \sqrt{17}}{2}$⁠, about $-1.56$ and $2.56$⁠. Both lie outside $[0, 2]$ (the figure shows the one at 2.56), so the fundamental theorem can be used. The bound shows this without decimals.
   Half the coefficient of $x$ is $-\frac{1}{2}$⁠; its square $\frac{1}{4}$ is added and subtracted. Basics: completing the square

2. Substitute $t = x - \frac{1}{2}$ and change the limits

   Then $dt = dx$⁠. Write $a = \frac{\sqrt{17}}{2}$⁠, so $a^2 = \frac{17}{4}$ and the denominator is $a^2 - t^2$⁠.

   New limits. When $x = 0$⁠, $t = -\frac{1}{2}$⁠. When $x = 2$⁠, $t = \frac{3}{2}$⁠.

   $$\int_0^2\frac{dx}{x + 4 - x^2} = \int_{-\frac{1}{2}}^{\frac{3}{2}}\frac{dt}{a^2 - t^2}.$$

   Why it is valid on the whole interval. $t = x - \frac{1}{2}$ is a shift: increasing, with derivative 1. Every $t$ in $\left[-\frac{1}{2}, \frac{3}{2}\right]$ has $|t| \le \frac{3}{2} < a \approx 2.06$⁠, so $a^2 - t^2 > 0$ throughout.

   After completing the square the expression depends on $x$ only through $x - \frac{1}{2}$⁠; naming it $t$ gives exactly the shape of a standard formula.
   $\sqrt{17} \approx 4.1231056$⁠, so $a \approx 2.0615528$⁠. Basics: substitution

3. Apply the formula for $\frac{1}{a^2 - t^2}$

   With $\frac{1}{2a} = \frac{1}{\sqrt{17}}$⁠:

   $$\int\frac{dt}{a^2 - t^2} = \frac{1}{\sqrt{17}}\log\left|\frac{a + t}{a - t}\right| + C.$$

   Since $|t| < a$ on our interval, both $a + t$ and $a - t$ are positive and the modulus can be dropped.

   Section 7.4 formula (2) comes from splitting $\frac{1}{a^2 - t^2} = \frac{1}{2a}\left(\frac{1}{a - t} + \frac{1}{a + t}\right)$ and integrating each part to a logarithm.
   Section 7.4, formula (2). Lesson: integrals of particular functions

4. Evaluate at the two limits

   Multiply top and bottom of $\frac{a + t}{a - t}$ by 2, so $a$ becomes $\sqrt{17}$ and $t$ becomes $2t$⁠:

   At $t = \frac{3}{2}$⁠: $\frac{\sqrt{17} + 3}{\sqrt{17} - 3}$⁠. At $t = -\frac{1}{2}$⁠: $\frac{\sqrt{17} - 1}{\sqrt{17} + 1}$⁠.

   $$\begin{aligned}I &= \frac{1}{\sqrt{17}}\left[\log\frac{\sqrt{17} + 3}{\sqrt{17} - 3} - \log\frac{\sqrt{17} - 1}{\sqrt{17} + 1}\right] \\ &= \frac{1}{\sqrt{17}}\left[\log\frac{\sqrt{17} + 3}{\sqrt{17} - 3} + \log\frac{\sqrt{17} + 1}{\sqrt{17} - 1}\right].\end{aligned}$$
   Theorem 2: value at the upper limit minus value at the lower limit. Subtracting $\log\frac{A}{B}$ is adding $\log\frac{B}{A}$⁠.
   $-\log\frac{A}{B} = \log\frac{B}{A}$⁠. Basics: logarithms

5. Simplify to the book's form

   Rationalise each fraction:

   $$\begin{aligned}\frac{\sqrt{17} + 3}{\sqrt{17} - 3} &= \frac{(\sqrt{17} + 3)^2}{17 - 9} = \frac{26 + 6\sqrt{17}}{8} \\ &= \frac{13 + 3\sqrt{17}}{4},\end{aligned}$$ $$\begin{aligned}\frac{\sqrt{17} + 1}{\sqrt{17} - 1} &= \frac{(\sqrt{17} + 1)^2}{17 - 1} = \frac{18 + 2\sqrt{17}}{16} \\ &= \frac{9 + \sqrt{17}}{8}.\end{aligned}$$

   Add the logarithms by multiplying:

   $$\begin{aligned}\frac{13 + 3\sqrt{17}}{4}\cdot\frac{9 + \sqrt{17}}{8} &= \frac{117 + 13\sqrt{17} + 27\sqrt{17} + 51}{32} \\ &= \frac{168 + 40\sqrt{17}}{32} = \frac{21 + 5\sqrt{17}}{4}.\end{aligned}$$

   Therefore

   $$I = \frac{1}{\sqrt{17}}\log\frac{21 + 5\sqrt{17}}{4}.$$

   In decimals: $\frac{21 + 5\sqrt{17}}{4} \approx 10.4038820$⁠, its logarithm is about 2.342179009, and dividing by $\sqrt{17} \approx 4.123105626$ gives $I \approx 0.5680618$⁠. (Nine decimals are kept here because 2.3421790 divided by 4.1231056 would round to 0.5680619.)

   $(\sqrt{17} + 3)^2 = 17 + 6\sqrt{17} + 9$ and $(\sqrt{17} + 1)^2 = 17 + 2\sqrt{17} + 1$⁠. In the product, $3\sqrt{17}\cdot\sqrt{17} = 51$⁠. Dividing $168 + 40\sqrt{17}$ and 32 by 8 gives $\frac{21 + 5\sqrt{17}}{4}$⁠.
   $\log A + \log B = \log(AB)$⁠; $(p + q)^2 = p^2 + 2pq + q^2$⁠. Basics: logarithms

6. Same result by staying in $x$

   With $t = x - \frac{1}{2}$⁠, $\frac{a + t}{a - t} = \frac{\sqrt{17} - 1 + 2x}{\sqrt{17} + 1 - 2x}$⁠. So an antiderivative on $[0, 2]$ is

   $$F(x) = \frac{1}{\sqrt{17}}\log\frac{\sqrt{17} - 1 + 2x}{\sqrt{17} + 1 - 2x}.$$

   Check: differentiate. Split the logarithm of the quotient:

   $$\begin{aligned}F(x) = \tfrac{1}{\sqrt{17}}\big[&\log(\sqrt{17} - 1 + 2x) \\ &- \log(\sqrt{17} + 1 - 2x)\big].\end{aligned}$$

   Each logarithm differentiates to (derivative of the inside) over (the inside):

   $$\begin{aligned}F'(x) &= \frac{1}{\sqrt{17}}\left[\frac{2}{\sqrt{17} - 1 + 2x} + \frac{2}{\sqrt{17} + 1 - 2x}\right] \\ &= \frac{1}{\sqrt{17}}\cdot\frac{2\cdot 2\sqrt{17}}{17 - (2x - 1)^2} \\ &= \frac{4}{17 - (4x^2 - 4x + 1)} \\ &= \frac{4}{16 + 4x - 4x^2} = \frac{1}{x + 4 - x^2},\end{aligned}$$

   the integrand.

   The two denominators are $\sqrt{17} + (2x - 1)$ and $\sqrt{17} - (2x - 1)$⁠; their sum is $2\sqrt{17}$ and their product is $17 - (2x - 1)^2$⁠.
   $\frac{d}{dx}\log(g(x)) = \frac{g'(x)}{g(x)}$ for $g(x) > 0$⁠. Basics: the chain rule backwards

7. The region

   The integrand stays between $\frac{4}{17} \approx 0.235$ and $\frac{1}{2}$ on an interval of length 2, so the area is between 0.47 and 1. The value 0.568 fits.
   For a function that is never negative, the definite integral is the area under its graph. Basics: area as a sum of thin strips

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