Partial fractions — linear, repeated, irreducible quadratic factors: Worked examples

Before anything, one promise: I will not use a symbol without saying what it means, and every step gets a "Why?". So:

Definition The words we will use over and over

$P (x), Q (x)$ ::: two polynomials (sums of powers of $x$ like $3 x^{2} - x + 1$ ). We integrate the fraction $\frac{P ( x )}{Q ( x )}$ .
$de g$ ::: the degree — the highest power of $x$ . $de g (3 x^{2} - x) = 2$ .
proper ::: $de g P < de g Q$ . Only then can we split directly (see parent).
irreducible quadratic ::: a quadratic $x^{2} + b x + c$ that has no real roots, so it cannot be broken into two linear factors. We test with the discriminant $b^{2} - 4 c$ : if it is negative, no real roots exist.
$ln$ ::: the natural logarithm, the function whose slope at $x$ is $1/ x$ . This is why $\int \frac{1}{x} d x = ln ∣ x ∣$ . The absolute value $∣ \cdot ∣$ is there because $ln$ only accepts positive inputs, but $x$ may be negative — so we always feed it the size of $x$ .
$arctan$ ::: the inverse of $tan$ : it answers "which angle has this tangent?" We meet it because $\int \frac{1}{x ^{2} + 1} d x = arctan x$ .

The scenario matrix

Every partial-fraction integral you will ever meet lands in one of these cells. Each example below is tagged with the cell it covers.

Cell	What makes it special	Example
A distinct linear	all factors linear, all different	Ex 1
B repeated linear	a factor like $(x - a)^{2}$	Ex 2
C irreducible quadratic → arctan	leftover constant over a quadratic	Ex 3
D quadratic → log + arctan	numerator forces both pieces	Ex 4
E improper (degenerate degree)	$de g P \geq de g Q$ : must divide first	Ex 5
F repeated quadratic	$(x^{2} + c)^{2}$ in the denominator	Ex 6
G "fake" quadratic (sign trap)	quadratic that secretly factors	Ex 7
H word problem	rate/mixing model, real units	Ex 8
I exam twist	Laplace inverse via partial fractions	Ex 9
J mixed (linear + quadratic)	both templates in one problem	Ex 10

Cell A — distinct linear factors

Worked example (1)

$\int \frac{5 x - 4}{x ^{2} - x - 6} d x$ Forecast: guess now — will the answer be logs, arctans, or a mix? (It should be pure logs: linear factors only.)

1. Factor the denominator. Why this step? We must know the pieces before we can un-add them. $x^{2} - x - 6 = (x - 3) (x + 2) .$ Proper? $de g 1 < de g 2$ ✓.

2. Write the template. Why? Each distinct linear factor gets one constant on top ([!formula] parent table). $\frac{5 x - 4}{( x - 3 ) ( x + 2 )} = \frac{A}{x - 3} + \frac{B}{x + 2} .$

3. Cover-up for $A$ . Why? Multiplying by $(x - 3)$ and setting $x = 3$ kills the $B$ term, which still carries $(x - 3) \to 0$ . $A = \frac{5 ( 3 ) - 4}{3 + 2} = \frac{11}{5} .$

4. Cover-up for $B$ . Why the same trick at $x = - 2$ ? Multiplying by $(x + 2)$ and setting $x = - 2$ kills the $A$ term (it carries $(x + 2) \to 0$ ), leaving $B$ alone. $B = \frac{5 ( - 2 ) - 4}{- 2 - 3} = \frac{- 14}{- 5} = \frac{14}{5} .$

5. Integrate each piece. Why log? $\int \frac{A}{x - a} d x = A ln ∣ x - a ∣$ . $\int = \frac{11}{5} ln ∣ x - 3∣ + \frac{14}{5} ln ∣ x + 2∣ + C .$

Verify: recombine — $\frac{11/5}{x - 3} + \frac{14/5}{x + 2} = \frac{\frac{11}{5} ( x + 2 ) + \frac{14}{5} ( x - 3 )}{( x - 3 ) ( x + 2 )} = \frac{5 x - 4}{x ^{2} - x - 6}$ ✓ (checked in VERIFY).

Cell B — repeated linear factor

Worked example (2)

$\int \frac{2 x - 1}{x ^{2} ( x - 3 )} d x$ Here $x^{2} = (x - 0)^{2}$ is a repeated linear factor.

Forecast: how many unknowns? (Count: $de g Q = 3$ , so exactly three.)

1. Full template — every power down. Why? One term $\frac{A _{2}}{x ^{2}}$ can only supply a constant numerator; we need the $\frac{A _{1}}{x}$ term to reach the linear freedom (parent intuition box). $\frac{2 x - 1}{x ^{2} ( x - 3 )} = \frac{A _{1}}{x} + \frac{A _{2}}{x ^{2}} + \frac{B}{x - 3} .$

2. Cover-up for $A_{2}$ (top power). Why it works only for the top power: multiplying by $x^{2}$ leaves $A_{2}$ bare at $x = 0$ . $A_{2} = \frac{2 ( 0 ) - 1}{0 - 3} = \frac{- 1}{- 3} = \frac{1}{3} .$

3. Cover-up for $B$ . Why? Multiplying by $(x - 3)$ and setting $x = 3$ kills both $A$ terms (each keeps an $x^{2}$ that stays finite while their denominators blow up — more directly, multiplying by $(x - 3)$ leaves the $A$ terms carrying $(x - 3) \to 0$ ), so only $B$ survives. $B = \frac{2 ( 3 ) - 1}{3 ^{2}} = \frac{5}{9} .$

4. Find $A_{1}$ by coefficient matching. Why? $A_{1}$ has no clean cover-up. Clear denominators: $2 x - 1 = A_{1} x (x - 3) + A_{2} (x - 3) + B x^{2} .$ Match the $x^{2}$ coefficient (fastest — only $A_{1}$ and $B$ live there): $0 = A_{1} + B \Rightarrow A_{1} = - \frac{5}{9} .$

5. Integrate. Why no log for the $x^{2}$ piece? $\int x^{- 2} d x = - x^{- 1}$ (power rule; exponent isn't $- 1$ ). $\int = - \frac{5}{9} ln ∣ x ∣ - \frac{1}{3} \cdot \frac{1}{x} + \frac{5}{9} ln ∣ x - 3∣ + C .$

Verify: at $x = 1$ : LHS integrand $\frac{2 - 1}{1 \cdot ( - 2 )} = - \frac{1}{2}$ ; pieces $\frac{- 5/9}{1} + \frac{1/3}{1} + \frac{5/9}{- 2} = - \frac{1}{2}$ ✓ (VERIFY).

Cell C — irreducible quadratic giving a pure arctan

Figure — Partial fractions — linear, repeated, irreducible quadratic factors

Figure s01 — Alt-text: the red curve is the integrand $\frac{1}{x ^{2} + 4}$ , a smooth symmetric bump peaking at $x = 0$ . The black curve is the running area swept out from the far left; it climbs steeply where the bump is tall and flattens out where the bump is small, tracing an S. That S is exactly the shape of $arctan$ — which is our clue that the antiderivative is an arctan and nothing else.

Worked example (3)

$\int \frac{4}{x ^{2} + 4} d x$ This is the degenerate quadratic case: numerator is a bare constant, no $x$ term.

Forecast: log or arctan? (No $x$ on top means no derivative-of-denominator to log — so pure arctan.)

1. Confirm irreducibility. Discriminant of $x^{2} + 0 x + 4$ is $0 - 16 = - 16 < 0$ ✓ — no real roots, cannot factor.

2. Match the standard arctan shape. Why this tool and not another? The only elementary function whose derivative is $\frac{1}{x ^{2} + d ^{2}}$ is $\frac{1}{d} arctan \frac{x}{d}$ . In s01 the red bump is the integrand and the black S-curve is the accumulated area: where the bump is tall the area climbs fast, where the bump fades the area levels off — read the S left-to-right and you are literally watching $arctan$ being built. See Standard integrals — log and arctan. $\int \frac{1}{x ^{2} + d ^{2}} d x = \frac{1}{d} arctan \frac{x}{d}, d = 2.$

3. Apply with the constant 4 pulled out. $\int \frac{4}{x ^{2} + 4} d x = 4 \cdot \frac{1}{2} arctan \frac{x}{2} + C = 2 arctan \frac{x}{2} + C .$

Verify: differentiate $2 arctan \frac{x}{2}$ : $2 \cdot \frac{1/2}{1 + ( x /2 ) ^{2}} = \frac{1}{1 + x ^{2} /4} = \frac{4}{4 + x ^{2}}$ ✓ (VERIFY).

Cell D — quadratic giving log and arctan

Figure s02 — Alt-text: two parabolas. The black one is $x^{2} + 6 x + 13$ with its lowest point (vertex) sitting at $x = - 3$ . The red one is $(x + 3)^{2} + 4$ : the very same parabola after we slide it 3 units right so its vertex sits on the vertical axis. Completing the square is exactly this slide; once the vertex is on the axis the denominator reads as $(shifted x)^{2} + 4$ , the clean $u^{2} + d^{2}$ form the arctan formula needs.

Worked example (4)

$\int \frac{3 x + 1}{x ^{2} + 6 x + 13} d x$ Numerator has an $x$ , so both pieces appear. This is the "log the derivative, arctan the square" move.

Forecast: we expect $(const) ln (\dots) + (const) arctan (\dots)$ .

1. Derivative of the denominator. Why? If the top were exactly $2 x + 6$ , the integral is instantly $ln$ (a substitution $u = x^{2} + 6 x + 13$ , see Integration by substitution). So we build that piece. $\frac{d}{d x} (x^{2} + 6 x + 13) = 2 x + 6.$

2. Rewrite the numerator around it. Why? Split the top into "a multiple of $2 x + 6$ " plus "a pure constant". Since $3 x + 1 = \frac{3}{2} (2 x + 6) + (1 - 9) = \frac{3}{2} (2 x + 6) - 8$ : $\frac{3 x + 1}{x ^{2} + 6 x + 13} = \frac{3}{2} \cdot \frac{2 x + 6}{x ^{2} + 6 x + 13} - \frac{8}{x ^{2} + 6 x + 13} .$

3. Log part. Why log here? Numerator is now the exact derivative of the denominator, so $\int \frac{u ^{'}}{u} d x = ln ∣ u ∣$ . $\int \frac{3}{2} \cdot \frac{2 x + 6}{x ^{2} + 6 x + 13} d x = \frac{3}{2} ln x^{2} + 6 x + 13 .$ (Here $x^{2} + 6 x + 13 = (x + 3)^{2} + 4 > 0$ always, so the bars are never strictly needed — but we keep the $∣ \cdot ∣$ so the convention "logs always carry absolute-value bars" stays uniform with the linear cases.)

4. Arctan part — complete the square. Why? To force the $\frac{1}{u ^{2} + d ^{2}}$ shape the arctan formula needs. In s02 the shift $x \to x + 3$ slides the parabola's vertex onto the axis, turning $x^{2} + 6 x + 13$ into $(x + 3)^{2} + 4$ (see Completing the square). $x^{2} + 6 x + 13 = (x + 3)^{2} + 4, d = 2.$ $- 8 \int \frac{d x}{( x + 3 ) ^{2} + 4} = - 8 \cdot \frac{1}{2} arctan \frac{x + 3}{2} = - 4 arctan \frac{x + 3}{2} .$

5. Combine. $\int = \frac{3}{2} ln x^{2} + 6 x + 13 - 4 arctan \frac{x + 3}{2} + C .$

Verify: differentiate the answer and simplify — it returns $\frac{3 x + 1}{x ^{2} + 6 x + 13}$ ✓ (VERIFY).

Cell E — improper (degree trap)

Worked example (5)

$\int \frac{x ^{3} + 2 x}{x ^{2} - 1} d x$ Degenerate degree: $de g P = 3 \geq de g Q = 2$ . Applying the template directly is the classic mistake.

Forecast: we must divide first, so expect a polynomial term plus logs.

1. Long-divide. Why? No sum of proper fractions can equal an improper function; their degree is always smaller. Use Polynomial long division. $\frac{x ^{3} + 2 x}{x ^{2} - 1} = x + \frac{3 x}{x ^{2} - 1} .$ (Check: $x (x^{2} - 1) + 3 x = x^{3} - x + 3 x = x^{3} + 2 x$ ✓.)

2. Partial-fraction the proper remainder. Why now and not before? Only after dividing is the leftover $\frac{3 x}{x ^{2} - 1}$ proper ( $de g 1 < de g 2$ ), so only now is the template legal. Factor $x^{2} - 1 = (x - 1) (x + 1)$ : $\frac{3 x}{( x - 1 ) ( x + 1 )} = \frac{A}{x - 1} + \frac{B}{x + 1} .$ Cover-up: $A = \frac{3 ( 1 )}{1 + 1} = \frac{3}{2}$ , $B = \frac{3 ( - 1 )}{- 1 - 1} = \frac{3}{2}$ .

3. Integrate all parts. Why? The polynomial $x$ integrates by the power rule to $\frac{x ^{2}}{2}$ ; each linear piece integrates to a log. $\int = \frac{x ^{2}}{2} + \frac{3}{2} ln ∣ x - 1∣ + \frac{3}{2} ln ∣ x + 1∣ + C .$

Verify: differentiate: $x + \frac{3}{2} (\frac{1}{x - 1} + \frac{1}{x + 1}) = x + \frac{3}{2} \cdot \frac{2 x}{x ^{2} - 1} = x + \frac{3 x}{x ^{2} - 1} = \frac{x ^{3} + 2 x}{x ^{2} - 1}$ ✓ (VERIFY).

Cell F — repeated irreducible quadratic

Worked example (6)

$\int \frac{x ^{2} + 2}{( x ^{2} + 1 ) ^{2}} d x$ A quadratic raised to a power. Template needs a $B x + C$ block for each power.

Forecast: likely an $arctan$ plus an algebraic (non-log) fraction.

1. Full quadratic template. Why every power? Same logic as repeated linear — a single block can't span all numerators. $\frac{x ^{2} + 2}{( x ^{2} + 1 ) ^{2}} = \frac{B _{1} x + C _{1}}{x ^{2} + 1} + \frac{B _{2} x + C _{2}}{( x ^{2} + 1 ) ^{2}} .$

2. Clear denominators and match. Why matching, not cover-up? Cover-up needs a real root to plug in; $x^{2} + 1$ has none, so we compare coefficients instead. Multiply by $(x^{2} + 1)^{2}$ : $x^{2} + 2 = (B_{1} x + C_{1}) (x^{2} + 1) + (B_{2} x + C_{2}) .$ Expand RHS $= B_{1} x^{3} + C_{1} x^{2} + (B_{1} + B_{2}) x + (C_{1} + C_{2})$ . Match:

$x^{3} : B_{1} = 0$
$x^{2} : C_{1} = 1$
$x^{1} : B_{1} + B_{2} = 0 \Rightarrow B_{2} = 0$
$x^{0} : C_{1} + C_{2} = 2 \Rightarrow C_{2} = 1$ $\frac{x ^{2} + 2}{( x ^{2} + 1 ) ^{2}} = \frac{1}{x ^{2} + 1} + \frac{1}{( x ^{2} + 1 ) ^{2}} .$

3. First piece → arctan. Why? It is the standard shape with $d = 1$ : $\int \frac{1}{x ^{2} + 1} d x = arctan x$ .

4. Second piece — derive the reduction formula. Why derive, not just quote? So you can trust it. We want $I = \int \frac{d x}{( x ^{2} + 1 ) ^{2}}$ . Start from the known $\int \frac{d x}{x ^{2} + 1} = arctan x$ and differentiate the trial piece $\frac{x}{x ^{2} + 1}$ by the quotient rule: $\frac{d}{d x} (\frac{x}{x ^{2} + 1}) = \frac{( x ^{2} + 1 ) - x ( 2 x )}{( x ^{2} + 1 ) ^{2}} = \frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}} .$ Now write $1 - x^{2} = 2 - (x^{2} + 1)$ , so $\frac{d}{d x} (\frac{x}{x ^{2} + 1}) = \frac{2}{( x ^{2} + 1 ) ^{2}} - \frac{1}{x ^{2} + 1} .$ Integrate both sides (LHS integrates to the bracket, and $\int \frac{1}{x ^{2} + 1} = arctan x$ ): $\frac{x}{x ^{2} + 1} = 2 I - arctan x \Rightarrow I = \frac{1}{2} (\frac{x}{x ^{2} + 1} + arctan x) .$

5. Combine. $\int = arctan x + \frac{1}{2} \cdot \frac{x}{x ^{2} + 1} + \frac{1}{2} arctan x + C = \frac{3}{2} arctan x + \frac{x}{2 ( x ^{2} + 1 )} + C .$

Verify: differentiate — returns $\frac{x ^{2} + 2}{( x ^{2} + 1 ) ^{2}}$ ✓ (VERIFY).

Cell G — the "fake quadratic" sign trap

Worked example (7)

$\int \frac{7}{x ^{2} - 2 x - 3} d x$ $x^{2} - 2 x - 3$ looks like it wants a $\frac{B x + C}{\dots}$ block. Trap!

Forecast: check the discriminant before deciding.

1. Discriminant test. Why first? Only truly irreducible quadratics get $B x + C$ ; if it factors we must use two linear blocks instead (see Factoring polynomials over the reals). $(- 2)^{2} - 4 (1) (- 3) = 4 + 12 = 16 > 0.$ Positive ⇒ real roots exist ⇒ it factors: $(x - 3) (x + 1)$ .

2. Treat as two linear factors. Why cover-up now works: the denominator has genuine real roots $x = 3$ and $x = - 1$ to plug in, exactly what cover-up needs. $\frac{7}{( x - 3 ) ( x + 1 )} = \frac{A}{x - 3} + \frac{B}{x + 1}, A = \frac{7}{3 + 1} = \frac{7}{4}, B = \frac{7}{- 1 - 3} = - \frac{7}{4} .$

3. Integrate. Why? Each linear piece is a log; the difference of logs collapses to a single log of a ratio. $\int = \frac{7}{4} ln ∣ x - 3∣ - \frac{7}{4} ln ∣ x + 1∣ + C = \frac{7}{4} ln \frac{x - 3}{x + 1} + C .$

Verify: $\frac{7/4}{x - 3} - \frac{7/4}{x + 1} = \frac{7}{4} \cdot \frac{4}{( x - 3 ) ( x + 1 )} = \frac{7}{x ^{2} - 2 x - 3}$ ✓ (VERIFY).

Cell H — word problem (real units)

Worked example (8) A tank drains so that the outflow rate is

$\frac{d t}{d V} = \frac{1}{V ^{2} - 4 V + 3}$ seconds per litre, where $V$ is volume in litres. How long to drain from $V = 5$ L to $V = 4$ L?

Forecast: integrate a rational function between limits → expect a difference of logs, answer in seconds.

1. Set up the definite integral with the right limits. Why the limits are $4$ to $5$ , not $5$ to $4$ : the physical process runs from $V = 5$ down to $V = 4$ , so a naive read gives $\int_{5}^{4}$ . But $\int_{5}^{4} = - \int_{4}^{5}$ , and elapsed time must be positive. We fix the sign once, up front, by writing the total time as the positive accumulation $T = \int_{4}^{5} \frac{d V}{V ^{2} - 4 V + 3},$ i.e. we integrate "seconds per litre" over the one litre of volume that passes between the two levels, lower limit below the upper. (If instead you kept $\int_{5}^{4}$ you would get a negative number and simply drop the sign — same answer.)

2. Factor & split. Why? The denominator factors into real linear pieces, so partial fractions applies (discriminant $16 > 0$ ). $V^{2} - 4 V + 3 = (V - 1) (V - 3)$ : $\frac{1}{( V - 1 ) ( V - 3 )} = \frac{A}{V - 1} + \frac{B}{V - 3}, A = \frac{1}{1 - 3} = - \frac{1}{2}, B = \frac{1}{3 - 1} = \frac{1}{2} .$

3. Antiderivative. Why combine into one log? The two logs share a coefficient magnitude $\frac{1}{2}$ with opposite signs, so they merge into a single $ln$ of a ratio, which is cleaner to evaluate at the limits. $F (V) = - \frac{1}{2} ln ∣ V - 1∣ + \frac{1}{2} ln ∣ V - 3∣ = \frac{1}{2} ln \frac{V - 3}{V - 1} .$

4. Evaluate $F (5) - F (4)$ . Why $F (5) - F (4)$ ? By the Fundamental Theorem, a definite integral is upper-limit value minus lower-limit value. $F (5) = \frac{1}{2} ln \frac{2}{4} = \frac{1}{2} ln \frac{1}{2}, F (4) = \frac{1}{2} ln \frac{1}{3} .$ $T = \frac{1}{2} ln \frac{1}{2} - \frac{1}{2} ln \frac{1}{3} = \frac{1}{2} ln \frac{3}{2} \approx 0.2027 s .$

Verify: $\frac{1}{2} ln (1.5) = 0.20273 \dots$ seconds — positive time, correct units ✓ (VERIFY).

Cell I — exam twist: inverse Laplace via partial fractions

Worked example (9) Find

$f (t)$ if its Laplace transform is $F (s) = \frac{s + 3}{s ( s ^{2} + 2 s + 5 )}$ .

Forecast: the $s$ factor gives a constant, the quadratic gives an exponential-times-sinusoid (its roots are complex).

1. Partial-fraction in $s$ . Why? Standard inverse-Laplace tables only know simple pieces, so we must split first — the same template machinery, now with $s$ as the variable. $\frac{s + 3}{s ( s ^{2} + 2 s + 5 )} = \frac{A}{s} + \frac{B s + C}{s ^{2} + 2 s + 5} .$

2. Cover-up for $A$ . Why cover-up here: $s = 0$ is a real root of the denominator, so multiplying by $s$ and setting $s = 0$ kills the quadratic term (it keeps a factor $s \to 0$ ) and isolates $A$ . $A = \frac{0 + 3}{0 + 0 + 5} = \frac{3}{5} .$

3. Match for $B, C$ . Why matching: the quadratic has no real root to plug in, so we compare coefficients. Clear denominators: $s + 3 = A (s^{2} + 2 s + 5) + (B s + C) s .$

$s^{2} : 0 = A + B \Rightarrow B = - \frac{3}{5}$
$s^{1} : 1 = 2 A + C \Rightarrow C = 1 - \frac{6}{5} = - \frac{1}{5}$ $F (s) = \frac{3/5}{s} + \frac{- \frac{3}{5} s - \frac{1}{5}}{s ^{2} + 2 s + 5} .$

4. Complete the square in the quadratic: $s^{2} + 2 s + 5 = (s + 1)^{2} + 2^{2}$ . Why? The inverse-Laplace table is written in the shifted form $(s + a)^{2} + ω^{2}$ , which decodes to $e^{- a t}$ times a sinusoid of frequency $ω$ — so we must first rewrite the top around $(s + 1)$ : $- \frac{3}{5} s - \frac{1}{5} = - \frac{3}{5} (s + 1) + \frac{2}{5} .$ So the quadratic block is $- \frac{3}{5} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 4} + \frac{2}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 4} .$

5. Read off inverses. Why these? Table: $\frac{s + 1}{( s + 1 ) ^{2} + 4} \to e^{- t} cos 2 t$ , and $\frac{2}{( s + 1 ) ^{2} + 4} \to e^{- t} sin 2 t$ . $f (t) = \frac{3}{5} - \frac{3}{5} e^{- t} cos 2 t + \frac{2}{5} \cdot \frac{1}{2} e^{- t} sin 2 t = \frac{3}{5} - \frac{3}{5} e^{- t} cos 2 t + \frac{1}{5} e^{- t} sin 2 t .$

Verify: at $t = 0$ , $f (0) = \frac{3}{5} - \frac{3}{5} + 0 = 0$ , matching the initial-value theorem $lim_{s \to \infty} s F (s) = lim \frac{s ( s + 3 )}{s ( s ^{2} + 2 s + 5 )} = 0$ ✓ (VERIFY).

Cell J — mixed: linear and irreducible quadratic together

Worked example (10)

$\int \frac{2 x ^{2} - x + 4}{( x - 1 ) ( x ^{2} + 4 )} d x$ The most general case: one linear factor and one irreducible quadratic, so both templates appear at once and interact.

Forecast: expect a $ln ∣ x - 1∣$ (from the linear), a $ln (x^{2} + 4)$ (log part of the quadratic), and an $arctan$ (arctan part of the quadratic).

1. Confirm the quadratic is irreducible & write the joint template. Why? $x^{2} + 4$ has discriminant $- 16 < 0$ , so it stays whole and needs a linear top $B x + C$ ; the linear factor needs a lone constant $A$ . $\frac{2 x ^{2} - x + 4}{( x - 1 ) ( x ^{2} + 4 )} = \frac{A}{x - 1} + \frac{B x + C}{x ^{2} + 4} .$ Counting check: total unknowns $= 3 = de g Q$ ✓.

2. Cover-up for $A$ . Why cover-up works for the linear piece: $x = 1$ is a real root; multiplying by $(x - 1)$ and setting $x = 1$ kills the quadratic term. $A = \frac{2 ( 1 ) ^{2} - 1 + 4}{1 ^{2} + 4} = \frac{5}{5} = 1.$

3. Match for $B, C$ . Why matching, not cover-up: the quadratic has no real root to plug in, so we compare coefficients. Clear denominators with $A = 1$ : $2 x^{2} - x + 4 = (x^{2} + 4) + (B x + C) (x - 1) .$ Expand RHS $= x^{2} + 4 + B x^{2} - B x + C x - C = (1 + B) x^{2} + (C - B) x + (4 - C)$ . Match:

$x^{2} : 1 + B = 2 \Rightarrow B = 1$
$x^{0} : 4 - C = 4 \Rightarrow C = 0$
(check $x^{1} : C - B = - 1$ ✓) $\frac{2 x ^{2} - x + 4}{( x - 1 ) ( x ^{2} + 4 )} = \frac{1}{x - 1} + \frac{x}{x ^{2} + 4} .$

4. Integrate each piece. Why these forms? The linear gives a log; the $\frac{x}{x ^{2} + 4}$ has (a multiple of) the denominator's derivative $2 x$ on top, so it is a log by substitution $u = x^{2} + 4$ — no arctan needed here because $C = 0$ left no bare constant. $\int = ln ∣ x - 1∣ + \frac{1}{2} ln x^{2} + 4 + C .$ (Had $C$ been nonzero, a term $\frac{C}{x ^{2} + 4}$ would have added a $\frac{C}{2} arctan \frac{x}{2}$ — the full mixed answer would then carry all three pieces.)

Verify: differentiate: $\frac{1}{x - 1} + \frac{x}{x ^{2} + 4} = \frac{( x ^{2} + 4 ) + x ( x - 1 )}{( x - 1 ) ( x ^{2} + 4 )} = \frac{2 x ^{2} - x + 4}{( x - 1 ) ( x ^{2} + 4 )}$ ✓ (VERIFY).

Recall

Recall Which cell is which?

Pure constant over irreducible quadratic gives? ::: A pure $arctan$ (Cell C). Numerator has an $x$ over irreducible quadratic gives? ::: log plus arctan (Cell D). First thing to do if $de g P \geq de g Q$ ? ::: Polynomial long division (Cell E). A quadratic with positive discriminant should be treated as? ::: Two linear factors (Cell G). Repeated quadratic $(x^{2} + 1)^{2}$ template? ::: $\frac{B _{1} x + C _{1}}{x ^{2} + 1} + \frac{B _{2} x + C _{2}}{( x ^{2} + 1 ) ^{2}}$ (Cell F). A problem with one linear and one irreducible-quadratic factor needs? ::: both templates $\frac{A}{x - a} + \frac{B x + C}{x ^{2} + \dots}$ at once (Cell J).

Prerequisites worth a detour: Completing the square, Integration by substitution, Standard integrals — log and arctan, Factoring polynomials over the reals.