Exercise 7.8 · Q14: $\int_0^1 \frac{2x + 3}{5x^2 + 1}\,dx$

Basics you need

Solution

1. Check continuity on $[0, 1]$

   Numerator and denominator are polynomials, and $5x^2 + 1 \ge 1 > 0$ for every real $x$⁠. So the integrand is continuous on all of $\mathbb{R}$⁠, in particular on $[0, 1]$⁠.

   Theorem 2 needs continuity on the closed interval. A quotient of polynomials is continuous wherever its denominator is not 0, and this one never is.
   $5x^2 \ge 0$⁠, so $5x^2 + 1 \ge 1$⁠. Lesson: the fundamental theorem

2. Split the fraction

   $$\frac{2x + 3}{5x^2 + 1} = \frac{2x}{5x^2 + 1} + \frac{3}{5x^2 + 1}.$$
   The two pieces need different methods. The derivative of $5x^2 + 1$ is $10x$⁠, a multiple of $2x$⁠, so the first piece is a log; the second has a constant on top, so it is an inverse tangent.
   $\frac{p + q}{r} = \frac{p}{r} + \frac{q}{r}$⁠. Basics: splitting fractions

3. First piece: substitute $t = 5x^2 + 1$

   $dt = 10x\,dx$⁠, so $2x\,dx = \frac15\,dt$⁠. Then

   $$\begin{aligned}\int\frac{2x\,dx}{5x^2 + 1} &= \frac15\int\frac{dt}{t} = \frac15\log|t| \\ &= \frac15\log(5x^2 + 1).\end{aligned}$$
   $2x = \frac15(10x)$⁠, and $10x\,dx$ is exactly $dt$⁠. No absolute value is needed in the end because $5x^2 + 1 > 0$⁠.
   Standard formula (xii), Section 7.2. Lesson: integration by substitution

4. Second piece: formula 7.4 (3)

   Take 5 out of the denominator so that $x^2$ has coefficient 1:

   $$\frac{3}{5x^2 + 1} = \frac35\cdot\frac{1}{x^2 + \frac15} = \frac35\cdot\frac{1}{x^2 + \left(\frac{1}{\sqrt5}\right)^2}.$$

   With $a = \frac{1}{\sqrt5}$ we have $\frac1a = \sqrt5$ and $\frac{x}{a} = \sqrt5\,x$⁠, so

   $$\begin{aligned}\int\frac{3\,dx}{5x^2 + 1} &= \frac35\cdot\sqrt5\,\tan^{-1}(\sqrt5\,x) \\ &= \frac{3}{\sqrt5}\tan^{-1}(\sqrt5\,x).\end{aligned}$$

   Together, $F(x) = \frac15\log(5x^2 + 1) + \frac{3}{\sqrt5}\tan^{-1}(\sqrt5\,x)$⁠.

   Formula 7.4 (3) is stated for $x^2 + a^2$ with leading coefficient 1, so the 5 must come out first. $\frac35\sqrt5 = \frac{3\sqrt5}{5} = \frac{3}{\sqrt5}$⁠, since $\sqrt5\cdot\sqrt5 = 5$⁠.
   Formula (3), Section 7.4. Lesson: integrals of particular functions

5. Check: differentiate $F$ back

   $$\begin{aligned}F'(x) &= \frac15\cdot\frac{10x}{5x^2 + 1} + \frac{3}{\sqrt5}\cdot\frac{\sqrt5}{1 + 5x^2} \\ &= \frac{2x}{5x^2 + 1} + \frac{3}{5x^2 + 1} \\ &= \frac{2x + 3}{5x^2 + 1}.\ \checkmark\end{aligned}$$
   For the second term: $\frac{d}{dx}\tan^{-1}u = \frac{1}{1 + u^2}\cdot\frac{du}{dx}$ with $u = \sqrt5\,x$⁠, $u^2 = 5x^2$⁠, $\frac{du}{dx} = \sqrt5$⁠.
   Standard formula (x), Section 7.2, and the chain rule. Basics: integration undoes differentiation

6. Evaluate $F(1) - F(0)$

   At $x = 1$⁠: $5x^2 + 1 = 6$ and $\sqrt5\,x = \sqrt5$⁠. At $x = 0$⁠: $\log 1 = 0$ and $\tan^{-1}0 = 0$⁠. So

   $$\begin{aligned}I &= \left[\tfrac15\log 6 + \tfrac{3}{\sqrt5}\tan^{-1}\sqrt5\right] - [0 + 0] \\ &= \frac15\log 6 + \frac{3}{\sqrt5}\tan^{-1}\sqrt5.\end{aligned}$$

   In decimals: $\frac15\log 6 \approx 0.3583519$⁠, $\tan^{-1}\sqrt5 \approx 1.1502620$ (radians), $\frac{3}{\sqrt5} \approx 1.3416408$⁠, so $I \approx 0.3583519 + 1.5432384$⁠, that is $I \approx 1.9015903$⁠.

   The lower limit contributes nothing because both $\log 1$ and $\tan^{-1}0$ are 0. $\tan^{-1}\sqrt5$ is not a standard angle, so the exact answer keeps it as it is.
   $\log 1 = 0$⁠, $\tan^{-1}0 = 0$⁠. Basics: logarithms and 1/x

Every fact used, and where it comes from