Partial fractions — linear, repeated, irreducible quadratic factors: Exercises

Before we begin, three tools you will lean on. Each is a link to its own vault note — click if it feels new:

Polynomial long division — used to make an improper fraction proper.
Completing the square — turns $x^{2} + b x + c$ into $(x + \frac{b}{2})^{2} + d^{2}$ , the shape an arctan needs.
Integration by substitution — the engine behind $\int \frac{2 x + b}{x ^{2} + b x + c} d x = ln (x^{2} + b x + c)$ .
Factoring polynomials over the reals — tells you which template blocks appear.

Level 1 — Recognition

You are not solving yet. You are choosing the correct template: the skeleton of unknown numerators, with no values found. The rule (from the parent): each factor of the denominator contributes one block, and the total number of unknowns equals $de g Q$ .

Recall Solution — L1·Q1

Two distinct linear factors. Each linear factor $(x - a)$ contributes a single constant over itself: $\frac{5 x - 2}{( x - 3 ) ( x + 4 )} = \frac{A}{x - 3} + \frac{B}{x + 4} .$ Count check: $de g Q = 2$ , and we have $2$ unknowns $A, B$ ✓.

Recall Solution — L1·Q2

$(x + 1)^{3}$ is a repeated linear factor of order $3$ : we must run every power down from $3$ to $1$ (one term alone cannot represent a numerator that could secretly be quadratic in $(x + 1)$ ). Plus one block for the lone $(x - 5)$ : $\frac{7}{( x + 1 ) ^{3} ( x - 5 )} = \frac{A _{1}}{x + 1} + \frac{A _{2}}{( x + 1 ) ^{2}} + \frac{A _{3}}{( x + 1 ) ^{3}} + \frac{B}{x - 5} .$ Count check: $de g Q = 4$ , unknowns $A_{1}, A_{2}, A_{3}, B = 4$ ✓.

Recall Solution — L1·Q3

Irreducible? Discriminant of $x^{2} + 0 x + 9$ is $0^{2} - 4 (1) (9) = - 36 < 0$ , so no real roots — yes, irreducible. An irreducible quadratic gets a linear top $B x + C$ (numerator degree one less than denominator). Proper? $de g P = 3$ , $de g Q = 1 + 2 = 3$ . Since $de g P = de g Q$ it is improper — the template alone is not enough; you would divide first. But the template for the remainder is: $\frac{x ^{3} + 1}{( x - 2 ) ( x ^{2} + 9 )} = (constant from division) + \frac{A}{x - 2} + \frac{B x + C}{x ^{2} + 9} .$ Count check (proper part): $de g Q = 3$ , unknowns $A, B, C = 3$ ✓.

Level 2 — Application

Now find the constants and integrate. Turn the crank with cover-up where it is clean, and matching coefficients where it is not.

Recall Solution — L2·Q1

Template: $\frac{5 x - 2}{( x - 3 ) ( x + 4 )} = \frac{A}{x - 3} + \frac{B}{x + 4} .$ Cover-up for $A$ : cover $(x - 3)$ , set $x = 3$ : $A = \frac{5 ( 3 ) - 2}{3 + 4} = \frac{13}{7} .$ (the $B$ term carried $(x - 3) \to 0$ .) Cover-up for $B$ : cover $(x + 4)$ , set $x = - 4$ : $B = \frac{5 ( - 4 ) - 2}{- 4 - 3} = \frac{- 22}{- 7} = \frac{22}{7} .$ Integrate (each $\frac{A}{x - a} \to A ln ∣ x - a ∣$ ): $\int = \frac{13}{7} ln ∣ x - 3∣ + \frac{22}{7} ln ∣ x + 4∣ + C .$

Recall Solution — L2·Q2

$(x - 2)^{2}$ repeated ⇒ template $\frac{3 x - 1}{( x - 2 ) ^{2}} = \frac{A}{x - 2} + \frac{B}{( x - 2 ) ^{2}} .$ $B$ (cover-up on top power): set $x = 2$ : $B = 3 (2) - 1 = 5.$ $A$ (matching): clear denominators: $3 x - 1 = A (x - 2) + B$ . Coefficient of $x$ : $A = 3$ . (Check constant: $- 2 A + B = - 6 + 5 = - 1$ ✓.) Integrate: the second piece uses the power rule, not a log (exponent $- 2 \neq = - 1$ ): $\int = 3 ln ∣ x - 2∣ - \frac{5}{x - 2} + C .$

Recall Solution — L2·Q3

There is no linear factor to cover up — split by the parent rule "log the derivative, arctan the square." The derivative of $x^{2} + 4$ is $2 x$ , so engineer the numerator: $4 x + 1 = 2 \cdot (2 x) + 1.$ $\int \frac{4 x + 1}{x ^{2} + 4} d x = 2 \int \frac{2 x}{x ^{2} + 4} d x + \int \frac{1}{x ^{2} + 4} d x .$ First integral by substitution $u = x^{2} + 4$ : $2 ln (x^{2} + 4)$ . Second is a standard arctan with $d^{2} = 4$ , $d = 2$ : $\int \frac{1}{x ^{2} + 2 ^{2}} d x = \frac{1}{2} arctan \frac{x}{2}$ . $\int = 2 ln (x^{2} + 4) + \frac{1}{2} arctan \frac{x}{2} + C .$

Level 3 — Analysis

Here the choice of method matters. A blind approach still works but wastes minutes; the smart move is short.

Recall Solution — L3·Q1

Template: $\frac{A}{x - 1} + \frac{B}{( x - 1 ) ^{2}} + \frac{C}{x + 2} .$ Cover-up $B$ (top power of the repeat): $x = 1$ : $B = \frac{1 + 4}{1 + 2} = \frac{5}{3} .$ Cover-up $C$ : $x = - 2$ : $C = \frac{- 2 + 4}{( - 3 ) ^{2}} = \frac{2}{9} .$ $A$ — the analysis: $A$ has no clean cover-up. Multiplying out, the numerator's $x^{2}$ coefficient is the only place $A$ and $C$ meet without $B$ . Since the left numerator $x + 4$ has no $x^{2}$ term: $A + C = 0 \Rightarrow A = - C = - \frac{2}{9} .$ Why choose the $x^{2}$ coefficient? it gives $A$ in one line instead of solving a $3 \times 3$ system. Integrate: $\int = - \frac{2}{9} ln ∣ x - 1∣ - \frac{5}{3} \cdot \frac{1}{x - 1} + \frac{2}{9} ln ∣ x + 2∣ + C .$

Recall Solution — L3·Q2

Check irreducible: discriminant $2^{2} - 4 (5) = 4 - 20 = - 16 < 0$ ✓. Derivative of denominator is $2 x + 2$ . Engineer the numerator so the derivative appears exactly: $x + 3 = \frac{1}{2} (2 x + 2) + 2.$ $\int = \frac{1}{2} \int \frac{2 x + 2}{x ^{2} + 2 x + 5} d x + \int \frac{2}{x ^{2} + 2 x + 5} d x .$ First piece: $\frac{1}{2} ln (x^{2} + 2 x + 5)$ . Second: complete the square $x^{2} + 2 x + 5 = (x + 1)^{2} + 4$ , so $d = 2$ : $\int \frac{2}{( x + 1 ) ^{2} + 2 ^{2}} d x = 2 \cdot \frac{1}{2} arctan \frac{x + 1}{2} = arctan \frac{x + 1}{2} .$ $\int = \frac{1}{2} ln (x^{2} + 2 x + 5) + arctan \frac{x + 1}{2} + C .$

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

The figure shows why the split is honest: the numerator $x + 3$ (coral) is exactly the derivative-part (mint, height $\frac{1}{2}$ of slope $2 x + 2$ ) plus a flat leftover (butter, height $2$ ). No guessing — every bit of $x + 3$ is accounted for.

Level 4 — Synthesis

One problem, several tools chained together.

Recall Solution — L4·Q1

Template: $\frac{A}{x + 1} + \frac{B x + C}{x ^{2} + 1}$ (quadratic irreducible: discriminant $- 4 < 0$ ). Cover-up $A$ : $x = - 1$ : $A = \frac{2 ( 1 ) - 1 + 1}{( - 1 ) ^{2} + 1} = \frac{2}{2} = 1.$ $B, C$ by clearing: $2 x^{2} + x + 1 = A (x^{2} + 1) + (B x + C) (x + 1)$ . With $A = 1$ : $2 x^{2} + x + 1 = (1 + B) x^{2} + (B + C) x + (1 + C) .$ Match $x^{2} : 1 + B = 2 \Rightarrow B = 1$ . Match const: $1 + C = 1 \Rightarrow C = 0$ . Check $x$ : $B + C = 1$ ✓. So $\frac{2 x ^{2} + x + 1}{( x + 1 ) ( x ^{2} + 1 )} = \frac{1}{x + 1} + \frac{x}{x ^{2} + 1} .$ Integrate (log for the linear, half-log via $u = x^{2} + 1$ for the quadratic; no arctan since $C = 0$ ): $\int = ln ∣ x + 1∣ + \frac{1}{2} ln (x^{2} + 1) + C .$

Recall Solution — L4·Q2

Proper? $de g P = 2 = de g Q$ ⇒ improper. Long-divide first: $\frac{x ^{2}}{x ^{2} - 1} = 1 + \frac{1}{x ^{2} - 1} .$ Now $x^{2} - 1 = (x - 1) (x + 1)$ , both linear: $\frac{1}{( x - 1 ) ( x + 1 )} = \frac{A}{x - 1} + \frac{B}{x + 1} .$ Cover-up $A$ at $x = 1$ : $A = \frac{1}{1 + 1} = \frac{1}{2}$ . Cover-up $B$ at $x = - 1$ : $B = \frac{1}{- 1 - 1} = - \frac{1}{2}$ . Integrate (the constant $1$ integrates to $x$ ): $\int = x + \frac{1}{2} ln ∣ x - 1∣ - \frac{1}{2} ln ∣ x + 1∣ + C = x + \frac{1}{2} ln \frac{x - 1}{x + 1} + C .$

Level 5 — Mastery

Everything at once: repeated, quadratic, log and arctan in a single answer.

Recall Solution — L5·Q1

Irreducible? discriminant $2^{2} - 4 (2) = 4 - 8 = - 4 < 0$ ✓. Derivative of $x^{2} + 2 x + 2$ is $2 x + 2$ . Write $2 x + 5 = (2 x + 2) + 3$ . $\int = \int \frac{2 x + 2}{x ^{2} + 2 x + 2} d x + \int \frac{3}{x ^{2} + 2 x + 2} d x .$ First = $ln (x^{2} + 2 x + 2)$ . Second: complete the square $x^{2} + 2 x + 2 = (x + 1)^{2} + 1$ , $d = 1$ : $3 \int \frac{1}{( x + 1 ) ^{2} + 1} d x = 3 arctan (x + 1) .$ $\int = ln (x^{2} + 2 x + 2) + 3 arctan (x + 1) + C .$

Recall Solution — L5·Q2

Proper? $de g P = 3$ , $de g Q = 2 + 2 = 4$ , so $3 < 4$ ✓ proper. Template (repeated linear runs both powers; irreducible quadratic gets $B x + C$ ): $\frac{x ^{3} - 2 x + 3}{( x - 1 ) ^{2} ( x ^{2} + 1 )} = \frac{A}{x - 1} + \frac{B}{( x - 1 ) ^{2}} + \frac{C x + D}{x ^{2} + 1} .$ Count: $de g Q = 4$ , unknowns $A, B, C, D = 4$ ✓. Find $B$ (cover-up top power): $x = 1$ : $B = \frac{1 - 2 + 3}{1 ^{2} + 1} = \frac{2}{2} = 1.$ Clear denominators: $x^{3} - 2 x + 3 = A (x - 1) (x^{2} + 1) + B (x^{2} + 1) + (C x + D) (x - 1)^{2} .$ Expand with $B = 1$ and match powers:

$x^{3}$ : $A + C = 1$
$x^{2}$ : $- A + 1 - 2 C + D = 0$
$x^{1}$ : $A - 2 D + C = - 2$ … (using $A - 2 C + ...$ careful expansion below) Do it cleanly. $A (x - 1) (x^{2} + 1) = A (x^{3} - x^{2} + x - 1)$ . $(C x + D) (x - 1)^{2} = (C x + D) (x^{2} - 2 x + 1) = C x^{3} + (D - 2 C) x^{2} + (C - 2 D) x + D$ . Sum coefficients (add $B (x^{2} + 1) = x^{2} + 1$ ):
$x^{3} : A + C = 1$
$x^{2} : - A + (D - 2 C) + 1 = 0$
$x^{1} : A + (C - 2 D) = - 2$
$x^{0} : - A + D + 1 = 3$ From $x^{0}$ : $D = A + 2$ . Sub into $x^{3}$ and $x^{1}$ : $C = 1 - A$ , and $A + (1 - A) - 2 (A + 2) = - 2 \Rightarrow 1 - 2 A - 4 = - 2 \Rightarrow - 2 A = 1 \Rightarrow A = - \frac{1}{2} .$ Then $C = 1 - (- \frac{1}{2}) = \frac{3}{2}$ , $D = - \frac{1}{2} + 2 = \frac{3}{2}$ . (Check $x^{2}$ : $- (- \frac{1}{2}) + (\frac{3}{2} - 3) + 1 = \frac{1}{2} - \frac{3}{2} + 1 = 0$ ✓.) $\frac{x ^{3} - 2 x + 3}{( x - 1 ) ^{2} ( x ^{2} + 1 )} = \frac{- \frac{1}{2}}{x - 1} + \frac{1}{( x - 1 ) ^{2}} + \frac{\frac{3}{2} x + \frac{3}{2}}{x ^{2} + 1} .$

Active recall

Which fractions may be partial-fractioned directly?

Only proper ones,

de g P < de g Q

; else long-divide first.

Top-power coefficient of a repeated factor is found by?

Cover-up (multiply by the full power, set

x = a

Lower-power and quadratic

B x + C

coefficients are found by?

Matching coefficients / clearing denominators.

Integrating

\frac{B x + C}{x ^{2} + b x + c}

splits into which two shapes?

A log (of the derivative part) plus an arctan (of the completed square).

Total unknowns in any decomposition equals?

de g Q