Exercise 7.9 · Q7: $\int_{-1}^1\frac{dx}{x^2+2x+5}$

Basics you need

Solution

1. Complete the square and check the denominator

   $$x^2 + 2x + 5 = (x^2 + 2x + 1) + 4 = (x + 1)^2 + 2^2.$$

   A square is $\ge 0$⁠, so the denominator is at least 4 for every real $x$⁠. The integrand is therefore continuous on $[-1, 1]$⁠.

   Half the coefficient of $x$ is 1 and its square is 1; we borrow 1 from the 5. The form "square plus positive constant" shows at once that the denominator never vanishes and that the arctangent formula fits.
   $(x + 1)^2 = x^2 + 2x + 1$⁠. Basics: completing the square

2. Substitute $t = x + 1$ and change the limits

   Then $dt = dx$⁠.

   New limits. When $x = -1$⁠, $t = 0$⁠. When $x = 1$⁠, $t = 2$⁠.

   $$\int_{-1}^{1}\frac{dx}{(x + 1)^2 + 2^2} = \int_0^2\frac{dt}{t^2 + 2^2}.$$

   Why it is valid on the whole interval. $t = x + 1$ is a shift with derivative 1: it is increasing and carries $[-1, 1]$ onto $[0, 2]$⁠. The new integrand $\frac{1}{t^2 + 4}$ is continuous everywhere.

   The denominator depends on $x$ only through $x + 1$⁠, so naming it $t$ produces exactly the shape $t^2 + a^2$ with $a = 2$⁠.
   A shift does not change $dx$⁠: $\frac{d}{dx}(x + 1) = 1$⁠. Basics: substitution

3. Integrate in $t$

   $$\begin{aligned}\int_0^2\frac{dt}{t^2 + 2^2} &= \left[\frac{1}{2}\tan^{-1}\frac{t}{2}\right]_0^2 \\ &= \frac{1}{2}\tan^{-1}1 - \frac{1}{2}\tan^{-1}0 \\ &= \frac{1}{2}\cdot\frac{\pi}{4} - 0 = \frac{\pi}{8}.\end{aligned}$$

   In decimals, $\frac{\pi}{8} \approx 0.3926991$⁠.

   Section 7.4 formula (3) with $a = 2$⁠. At $t = 2$ the argument $\frac{t}{2}$ is 1, and $\tan^{-1}1 = \frac{\pi}{4}$ because $\frac{\pi}{4}$ is in the principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $\tan\frac{\pi}{4} = 1$⁠.
   Section 7.4, formula (3), then Theorem 2. Lesson: integrals of particular functions

4. Same result by staying in $x$

   An antiderivative in $x$ is

   $$F(x) = \frac{1}{2}\tan^{-1}\frac{x + 1}{2}.$$

   Check: differentiate. Chain rule with inner $u = \frac{x+1}{2}$⁠, $u' = \frac{1}{2}$⁠:

   $$\begin{aligned}F'(x) &= \frac{1}{2}\cdot\frac{1}{1 + \frac{(x+1)^2}{4}}\cdot\frac{1}{2} \\ &= \frac{1}{4 + (x + 1)^2} = \frac{1}{x^2 + 2x + 5},\end{aligned}$$

   the integrand. Then $F(1) - F(-1) = \frac{1}{2}\tan^{-1}1 - \frac{1}{2}\tan^{-1}0 = \frac{\pi}{8}$⁠.

   Multiplying $\frac{1}{4}$ into $\frac{1}{1 + (x+1)^2/4}$ clears the inner fraction: $4\left(1 + \frac{(x+1)^2}{4}\right) = 4 + (x+1)^2$⁠.
   $\frac{d}{du}\tan^{-1}u = \frac{1}{1+u^2}$ (Chapter 5, Section 5.3.3). Basics: the chain rule backwards

5. The region

   On $[-1, 1]$ the integrand falls from $\frac{1}{4}$ to $\frac{1}{8}$⁠, so the area lies between $2\times\frac{1}{8} = 0.25$ and $2\times\frac{1}{4} = 0.5$⁠. The value 0.393 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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