Exercise 7.8 · Q16: $\int_1^2 \frac{5x^2}{x^2 + 4x + 3}\,dx$

Basics you need

Solution

1. Factor the denominator and check continuity on $[1, 2]$

   $x^2 + 4x + 3 = (x + 1)(x + 3)$⁠, because $1 \cdot 3 = 3$ and $1 + 3 = 4$⁠. It is 0 only at $x = -1$ and $x = -3$⁠, both outside $[1, 2]$⁠. On $[1, 2]$ both factors are positive, so the integrand is continuous there.

   Theorem 2 needs continuity on the whole closed interval. A rational function is continuous wherever its denominator is not 0.
   Factorising a quadratic: find two numbers with product 3 and sum 4. Lesson: the fundamental theorem

2. Long division

   $5x^2 = 5(x^2 + 4x + 3) - 20x - 15$⁠, so

   $$\frac{5x^2}{x^2 + 4x + 3} = 5 - \frac{20x + 15}{(x + 1)(x + 3)}.$$
   Partial fractions only work for proper fractions. Dividing $5x^2$ by $x^2 + 4x + 3$ gives quotient 5; the remainder is $5x^2 - 5x^2 - 20x - 15 = -20x - 15$⁠, of degree 1, less than 2.
   $\frac{P}{Q} = T + \frac{P_1}{Q}$⁠, Section 7.5. Basics: splitting fractions

3. Partial fractions for $\frac{20x + 15}{(x + 1)(x + 3)}$

   Write $\frac{20x + 15}{(x + 1)(x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 3}$ and clear denominators:

   $$20x + 15 = A(x + 3) + B(x + 1).$$

   Put $x = -1$⁠: $-5 = 2A$⁠, so $A = -\frac52$⁠. Put $x = -3$⁠: $-45 = -2B$⁠, so $B = \frac{45}{2}$⁠.

   Coefficient check: $A + B = 20$ ✓ (coefficient of $x$⁠) and $3A + B = -\frac{15}{2} + \frac{45}{2} = 15$ ✓ (constant term). Therefore

   $$\begin{aligned}\frac{5x^2}{x^2 + 4x + 3} &= 5 - \frac{-\frac52}{x + 1} - \frac{\frac{45}{2}}{x + 3} \\ &= 5 + \frac{5}{2(x + 1)} - \frac{45}{2(x + 3)}.\end{aligned}$$
   The cleared identity holds for every $x$⁠, so we may put in the roots of the factors: each choice kills one constant and isolates the other. Watch the signs: subtracting $\frac{A}{x + 1}$ with $A = -\frac52$ gives $+\frac{5}{2(x + 1)}$⁠.
   Table 7.2, form 1, Section 7.5. Lesson: partial fractions

4. Integrate term by term

   $$F(x) = 5x + \frac52\log|x + 1| - \frac{45}{2}\log|x + 3|.$$

   On $[1, 2]$⁠, $x + 1$ and $x + 3$ are positive, so the bars can be dropped there.

   $\int 5\,dx = 5x$⁠, and $\int\frac{dx}{x + k} = \log|x + k|$ (shift $u = x + k$⁠, $du = dx$⁠, then standard formula (xii)). Constants come outside the integral.
   Standard formula (xii), Section 7.2; Properties (III), (IV), Section 7.2.1. Basics: logarithms and 1/x

5. Check: differentiate $F$ back

   $$F'(x) = 5 + \frac{5}{2(x + 1)} - \frac{45}{2(x + 3)}.$$

   Over the common denominator $2(x + 1)(x + 3)$ the numerator is

   $$\begin{aligned}&10(x^2 + 4x + 3) + 5(x + 3) - 45(x + 1) \\ &= 10x^2 + 40x + 30 + 5x + 15 - 45x - 45 \\ &= 10x^2,\end{aligned}$$

   so $F'(x) = \frac{10x^2}{2(x + 1)(x + 3)} = \frac{5x^2}{x^2 + 4x + 3}$⁠. ✓

   The $x$ terms cancel ($40 + 5 - 45 = 0$⁠) and so do the constants ($30 + 15 - 45 = 0$⁠), leaving exactly the original numerator.
   Property (I), Section 7.2.1. Basics: integration undoes differentiation

6. Evaluate $F(2) - F(1)$

   $$\begin{aligned}F(2) &= 10 + \tfrac52\log 3 - \tfrac{45}{2}\log 5, \\ F(1) &= 5 + \tfrac52\log 2 - \tfrac{45}{2}\log 4.\end{aligned}$$

   Subtract, grouping like logarithms:

   $$\begin{aligned}I &= 5 + \tfrac52(\log 3 - \log 2) - \tfrac{45}{2}(\log 5 - \log 4) \\ &= 5 + \frac52\log\frac32 - \frac{45}{2}\log\frac54.\end{aligned}$$

   In decimals: $\frac52\log\frac32 \approx 1.0136628$ and $\frac{45}{2}\log\frac54 \approx 5.0207299$⁠, so

   $$I \approx 5 + 1.0136628 - 5.0207299 = 0.9929329.$$

   The book's form. Take out $-\frac52$ from the two log terms, using $\frac{45}{2} = \frac52\cdot 9$⁠:

   $$I = 5 - \frac52\left(9\log\frac54 - \log\frac32\right),$$

   which is exactly the textbook answer.

   The large numbers 5 and 5.02 nearly cancel, which is why careful grouping matters: the answer is just under 1. The figure confirms the size.
   $\log a - \log b = \log\frac{a}{b}$⁠. Basics: logarithms and 1/x

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