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

4.2.10 · D3 · Maths › Calculus II — Integration › Partial fractions — linear, repeated, irreducible quadratic

Shuru karne se pehle, ek waada: main koi bhi symbol use nahi karunga bina bataye ki uska matlab kya hai, aur har step ke saath ek "Kyun?" milega. Toh:

Definition Woh words jo baar baar ayenge

$P (x), Q (x)$ ::: do polynomials ( $x$ ki powers ke sums jaise $3 x^{2} - x + 1$ ). Hum fraction $\frac{P ( x )}{Q ( x )}$ ko integrate karte hain.
$de g$ ::: degree — $x$ ki sabse badi power. $de g (3 x^{2} - x) = 2$ .
proper ::: $de g P < de g Q$ . Tabhi hum directly split kar sakte hain (parent dekho).
irreducible quadratic ::: ek quadratic $x^{2} + b x + c$ jiske real roots nahi hote, isliye ise do linear factors mein nahi toda ja sakta. Hum discriminant $b^{2} - 4 c$ se test karte hain: agar ye negative hai, toh koi real roots nahi hain.
$ln$ ::: natural logarithm, woh function jiska slope $x$ par $1/ x$ hai. Isliye $\int \frac{1}{x} d x = ln ∣ x ∣$ . Absolute value $∣ \cdot ∣$ isliye hai kyunki $ln$ sirf positive inputs leta hai, lekin $x$ negative ho sakta hai — toh hum hamesha $x$ ka size feed karte hain.
$arctan$ ::: $tan$ ka inverse: ye jawab deta hai "kis angle ka tangent ye hai?" Hum ise isliye milte hain kyunki $\int \frac{1}{x ^{2} + 1} d x = arctan x$ .

Scenario matrix

Har partial-fraction integral jo tum kabhi bhi miloge, in cells mein se kisi ek mein aata hai. Neeche har example ko us cell ke saath tag kiya gaya hai jo woh cover karta hai.

Cell	Kya khaas hai	Example
A distinct linear	saare factors linear, saare alag	Ex 1
B repeated linear	koi factor jaise $(x - a)^{2}$	Ex 2
C irreducible quadratic → arctan	quadratic ke upar sirf constant bacha	Ex 3
D quadratic → log + arctan	numerator dono pieces force karta hai	Ex 4
E improper (degenerate degree)	$de g P \geq de g Q$ : pehle divide karna hoga	Ex 5
F repeated quadratic	denominator mein $(x^{2} + c)^{2}$	Ex 6
G "fake" quadratic (sign trap)	quadratic jo secretly factor ho jaata hai	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)	ek hi problem mein dono templates	Ex 10

Cell A — distinct linear factors

Worked example (1)

$\int \frac{5 x - 4}{x ^{2} - x - 6} d x$ Forecast: abhi guess karo — answer logs hoga, arctans, ya mix? (Pure logs hona chahiye: sirf linear factors hain.)

1. Denominator ko factor karo. Kyun ye step? Pieces jaanne chahiye pehle unhe un-add karne se. $x^{2} - x - 6 = (x - 3) (x + 2) .$ Proper? $de g 1 < de g 2$ ✓.

2. Template likho. Kyun? Har distinct linear factor ke upar ek constant aata hai ([!formula] parent table). $\frac{5 x - 4}{( x - 3 ) ( x + 2 )} = \frac{A}{x - 3} + \frac{B}{x + 2} .$

3. Cover-up for $A$ . Kyun? $(x - 3)$ se multiply karke $x = 3$ set karne se $B$ term mar jaata hai, kyunki usme $(x - 3) \to 0$ hai. $A = \frac{5 ( 3 ) - 4}{3 + 2} = \frac{11}{5} .$

4. Cover-up for $B$ . Kyun same trick $x = - 2$ par? $(x + 2)$ se multiply karke $x = - 2$ set karne se $A$ term mar jaata hai (usme $(x + 2) \to 0$ hai), sirf $B$ bachta hai. $B = \frac{5 ( - 2 ) - 4}{- 2 - 3} = \frac{- 14}{- 5} = \frac{14}{5} .$

5. Har piece integrate karo. Kyun 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 karo — $\frac{11/5}{x - 3} + \frac{14/5}{x + 2} = \frac{\frac{11}{5} ( x + 2 ) + \frac{14}{5} ( x - 3 )}{( x - 3 ) ( x + 2 )} = \frac{5 x - 4}{x ^{2} - x - 6}$ ✓ (VERIFY mein check hua).

Cell B — repeated linear factor

Worked example (2)

$\int \frac{2 x - 1}{x ^{2} ( x - 3 )} d x$ Yahan $x^{2} = (x - 0)^{2}$ ek repeated linear factor hai.

Forecast: kitne unknowns honge? (Count karo: $de g Q = 3$ , toh exactly teen.)

1. Full template — har power neeche tak. Kyun? Ek term $\frac{A _{2}}{x ^{2}}$ sirf constant numerator de sakta hai; linear freedom ke liye hume $\frac{A _{1}}{x}$ term chahiye (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). Kyun sirf top power ke liye kaam karta hai: $x^{2}$ se multiply karne par $x = 0$ par $A_{2}$ akela bach jaata hai. $A_{2} = \frac{2 ( 0 ) - 1}{0 - 3} = \frac{- 1}{- 3} = \frac{1}{3} .$

3. Cover-up for $B$ . Kyun? $(x - 3)$ se multiply karke $x = 3$ set karne se dono $A$ terms mar jaate hain (cover-up logic se), sirf $B$ bachta hai. $B = \frac{2 ( 3 ) - 1}{3 ^{2}} = \frac{5}{9} .$

4. Coefficient matching se $A_{1}$ nikalo. Kyun? $A_{1}$ ke liye koi clean cover-up nahi hai. Denominators clear karo: $2 x - 1 = A_{1} x (x - 3) + A_{2} (x - 3) + B x^{2} .$ $x^{2}$ coefficient match karo (fastest — sirf $A_{1}$ aur $B$ wahan hain): $0 = A_{1} + B \Rightarrow A_{1} = - \frac{5}{9} .$

5. Integrate karo. Kyun $x^{2}$ piece ke liye log nahi? $\int x^{- 2} d x = - x^{- 1}$ (power rule; exponent $- 1$ nahi hai). $\int = - \frac{5}{9} ln ∣ x ∣ - \frac{1}{3} \cdot \frac{1}{x} + \frac{5}{9} ln ∣ x - 3∣ + C .$

Verify: $x = 1$ par: 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: laal curve integrand $\frac{1}{x ^{2} + 4}$ hai, ek smooth symmetric bump jo $x = 0$ par peak karta hai. Kaali curve woh running area hai jo bahut door left se sweep hoti hai; woh bump ke tall hone par teji se chadhti hai aur bump ke chote hone par flat ho jaati hai, ek S banate hue. Woh S exactly $arctan$ ki shape hai — jo hamara clue hai ki antiderivative arctan hai aur kuch nahi.

Worked example (3)

$\int \frac{4}{x ^{2} + 4} d x$ Ye degenerate quadratic case hai: numerator sirf ek bare constant hai, koi $x$ term nahi.

Forecast: log ya arctan? (Upar koi $x$ nahi matlab derivative-of-denominator log karne ke liye nahi — toh pure arctan.)

1. Irreducibility confirm karo. $x^{2} + 0 x + 4$ ka discriminant $0 - 16 = - 16 < 0$ ✓ — koi real roots nahi, factor nahi ho sakta.

2. Standard arctan shape se match karo. Kyun ye tool aur koi nahi? Woh akela elementary function jiska derivative $\frac{1}{x ^{2} + d ^{2}}$ hai woh hai $\frac{1}{d} arctan \frac{x}{d}$ . s01 mein laal bump integrand hai aur kaali S-curve accumulated area hai: jahan bump tall hai area fast chadhta hai, jahan bump fade hoti hai area level ho jaata hai — S ko left-to-right padho aur tum literally $arctan$ ko bante dekh rahe ho. Dekho Standard integrals — log and arctan. $\int \frac{1}{x ^{2} + d ^{2}} d x = \frac{1}{d} arctan \frac{x}{d}, d = 2.$

3. Constant 4 bahar khinch ke apply karo. $\int \frac{4}{x ^{2} + 4} d x = 4 \cdot \frac{1}{2} arctan \frac{x}{2} + C = 2 arctan \frac{x}{2} + C .$

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

Cell D — quadratic giving log and arctan

Figure s02 — Alt-text: do parabolas. Kaali $x^{2} + 6 x + 13$ hai jiska lowest point (vertex) $x = - 3$ par baitha hai. Laal $(x + 3)^{2} + 4$ hai: bilkul wohi parabola jise completing the square karke 3 units right slide kiya gaya hai taki uska vertex vertical axis par aa jaye. Completing the square exactly yahi slide hai; jab vertex axis par aa jaata hai tab denominator $(shifted x)^{2} + 4$ read hota hai, woh clean $u^{2} + d^{2}$ form jo arctan formula ko chahiye.

Worked example (4)

$\int \frac{3 x + 1}{x ^{2} + 6 x + 13} d x$ Numerator mein $x$ hai, toh dono pieces aate hain. Ye "derivative ko log karo, square ko arctan karo" wali move hai.

Forecast: hum expect karte hain $(const) ln (\dots) + (const) arctan (\dots)$ .

1. Denominator ka derivative. Kyun? Agar upar exactly $2 x + 6$ hota, toh integral instantly $ln$ hota (substitution $u = x^{2} + 6 x + 13$ , dekho Integration by substitution). Toh hum woh piece banate hain. $\frac{d}{d x} (x^{2} + 6 x + 13) = 2 x + 6.$

2. Numerator ko uske around rewrite karo. Kyun? Top ko " $2 x + 6$ ka ek multiple" aur "pure constant" mein split karo. Kyunki $3 x + 1 = \frac{3}{2} (2 x + 6) + (1 - 9) = \frac{3}{2} (2 x + 6) - 8$ : $\frac{3 x + 1}{x ^{2} + 6 x + 13} = \frac{3}{2} \cdot \frac{2 x + 6}{x ^{2} + 6 x + 13} - \frac{8}{x ^{2} + 6 x + 13} .$

3. Log part. Kyun yahan log? Numerator ab denominator ka exact derivative hai, toh $\int \frac{u ^{'}}{u} d x = ln ∣ u ∣$ . $\int \frac{3}{2} \cdot \frac{2 x + 6}{x ^{2} + 6 x + 13} d x = \frac{3}{2} ln x^{2} + 6 x + 13 .$ (Yahan $x^{2} + 6 x + 13 = (x + 3)^{2} + 4 > 0$ hamesha, toh bars strictly zaroori nahi — lekin hum $∣ \cdot ∣$ rakhte hain taaki convention "logs hamesha absolute-value bars lete hain" linear cases ke saath uniform rahe.)

4. Arctan part — complete the square karo. Kyun? $\frac{1}{u ^{2} + d ^{2}}$ shape force karne ke liye jo arctan formula ko chahiye. s02 mein shift $x \to x + 3$ parabola ke vertex ko axis par slide karta hai, $x^{2} + 6 x + 13$ ko $(x + 3)^{2} + 4$ mein badal deta hai (dekho 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 karo. $\int = \frac{3}{2} ln x^{2} + 6 x + 13 - 4 arctan \frac{x + 3}{2} + C .$

Verify: answer differentiate karo aur simplify karo — woh $\frac{3 x + 1}{x ^{2} + 6 x + 13}$ return karta hai ✓ (VERIFY).

Cell E — improper (degree trap)

Worked example (5)

$\int \frac{x ^{3} + 2 x}{x ^{2} - 1} d x$ Degenerate degree: $de g P = 3 \geq de g Q = 2$ . Template directly apply karna classic mistake hai.

Forecast: pehle divide karna hoga, toh ek polynomial term plus logs expect karo.

1. Long-divide karo. Kyun? Proper fractions ka koi bhi sum ek improper function ke barabar nahi ho sakta; unki degree hamesha chhoti hoti hai. Use karo 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. Proper remainder ka partial-fraction karo. Kyun ab aur pehle nahi? Sirf divide karne ke baad leftover $\frac{3 x}{x ^{2} - 1}$ proper hai ( $de g 1 < de g 2$ ), toh sirf ab template legal hai. 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. Saare parts integrate karo. Kyun? Polynomial $x$ power rule se $\frac{x ^{2}}{2}$ integrate hota hai; har linear piece ek log mein integrate hoti hai. $\int = \frac{x ^{2}}{2} + \frac{3}{2} ln ∣ x - 1∣ + \frac{3}{2} ln ∣ x + 1∣ + C .$

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

Cell F — repeated irreducible quadratic

Worked example (6)

$\int \frac{x ^{2} + 2}{( x ^{2} + 1 ) ^{2}} d x$ Ek quadratic power tak uthaya gaya. Template ko har power ke liye ek $B x + C$ block chahiye.

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

1. Full quadratic template. Kyun har power? Wohi logic jaise repeated linear — ek single block saare numerators span nahi kar sakta. $\frac{x ^{2} + 2}{( x ^{2} + 1 ) ^{2}} = \frac{B _{1} x + C _{1}}{x ^{2} + 1} + \frac{B _{2} x + C _{2}}{( x ^{2} + 1 ) ^{2}} .$

2. Denominators clear karo aur match karo. Kyun matching, cover-up nahi? Cover-up ko plug in karne ke liye real root chahiye; $x^{2} + 1$ mein koi nahi hai, toh hum coefficients compare karte hain. $(x^{2} + 1)^{2}$ se multiply karo: $x^{2} + 2 = (B_{1} x + C_{1}) (x^{2} + 1) + (B_{2} x + C_{2}) .$ RHS expand karo $= B_{1} x^{3} + C_{1} x^{2} + (B_{1} + B_{2}) x + (C_{1} + C_{2})$ . Match karo:

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

3. Pehla piece → arctan. Kyun? Ye standard shape hai $d = 1$ ke saath: $\int \frac{1}{x ^{2} + 1} d x = arctan x$ .

4. Doosra piece — reduction formula derive karo. Kyun derive karo, quote nahi? Taaki tum usp par trust kar sako. Hum chahte hain $I = \int \frac{d x}{( x ^{2} + 1 ) ^{2}}$ . Jaane-pahechaane $\int \frac{d x}{x ^{2} + 1} = arctan x$ se shuru karo aur trial piece $\frac{x}{x ^{2} + 1}$ ko quotient rule se differentiate karo: $\frac{d}{d x} (\frac{x}{x ^{2} + 1}) = \frac{( x ^{2} + 1 ) - x ( 2 x )}{( x ^{2} + 1 ) ^{2}} = \frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}} .$ Ab likho $1 - x^{2} = 2 - (x^{2} + 1)$ , toh $\frac{d}{d x} (\frac{x}{x ^{2} + 1}) = \frac{2}{( x ^{2} + 1 ) ^{2}} - \frac{1}{x ^{2} + 1} .$ Dono sides integrate karo (LHS bracket mein integrate hota hai, aur $\int \frac{1}{x ^{2} + 1} = arctan x$ ): $\frac{x}{x ^{2} + 1} = 2 I - arctan x \Rightarrow I = \frac{1}{2} (\frac{x}{x ^{2} + 1} + arctan x) .$

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

Verify: differentiate karo — $\frac{x ^{2} + 2}{( x ^{2} + 1 ) ^{2}}$ return karta hai ✓ (VERIFY).

Cell G — "fake quadratic" sign trap

Worked example (7)

$\int \frac{7}{x ^{2} - 2 x - 3} d x$ $x^{2} - 2 x - 3$ lagta hai jaise $\frac{B x + C}{\dots}$ block chahta hai. Trap!

Forecast: decide karne se pehle discriminant check karo.

1. Discriminant test. Kyun pehle? Sirf truly irreducible quadratics ko $B x + C$ milta hai; agar factor hota hai toh hume do linear blocks use karne chahiye (dekho Factoring polynomials over the reals). $(- 2)^{2} - 4 (1) (- 3) = 4 + 12 = 16 > 0.$ Positive ⇒ real roots exist hain ⇒ ye factors: $(x - 3) (x + 1)$ .

2. Do linear factors ki tarah treat karo. Kyun cover-up ab kaam karta hai: denominator ke genuine real roots $x = 3$ aur $x = - 1$ hain plug in karne ke liye, exactly wahi jo cover-up ko chahiye. $\frac{7}{( x - 3 ) ( x + 1 )} = \frac{A}{x - 3} + \frac{B}{x + 1}, A = \frac{7}{3 + 1} = \frac{7}{4}, B = \frac{7}{- 1 - 3} = - \frac{7}{4} .$

3. Integrate karo. Kyun? Har linear piece ek log hai; logs ka difference ek ratio ke single $ln$ mein collapse ho jaata hai. $\int = \frac{7}{4} ln ∣ x - 3∣ - \frac{7}{4} ln ∣ x + 1∣ + C = \frac{7}{4} ln \frac{x - 3}{x + 1} + C .$

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) Ek tank drain hota hai aur outflow rate hai

$\frac{d t}{d V} = \frac{1}{V ^{2} - 4 V + 3}$ seconds per litre, jahan $V$ litres mein volume hai. $V = 5$ L se $V = 4$ L tak drain hone mein kitna time lagega?

Forecast: limits ke beech ek rational function integrate karo → logs ka difference expect karo, answer seconds mein.

1. Sahi limits ke saath definite integral set up karo. Kyun limits $4$ se $5$ hain, $5$ se $4$ nahi: physical process $V = 5$ se $V = 4$ tak chalta hai, toh ek naive read $\int_{5}^{4}$ deta hai. Lekin $\int_{5}^{4} = - \int_{4}^{5}$ , aur elapsed time positive hona chahiye. Hum sign ek baar up front fix karte hain, total time ko positive accumulation ke roop mein likhke $T = \int_{4}^{5} \frac{d V}{V ^{2} - 4 V + 3},$ yaani hum "seconds per litre" ko volume ke ek litre par integrate karte hain jo dono levels ke beech pass hota hai, lower limit upper ke neeche. (Agar tum $\int_{5}^{4}$ rakhte toh negative number aata aur tum sirf sign drop karte — same answer.)

2. Factor aur split karo. Kyun? Denominator real linear pieces mein factor hota hai, toh partial fractions apply hota hai (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. Kyun ek log mein combine karo? Dono logs ka coefficient magnitude $\frac{1}{2}$ hai opposite signs ke saath, toh woh ek ratio ke single $ln$ mein merge ho jaate hain, jo limits par evaluate karne ke liye cleaner hai. $F (V) = - \frac{1}{2} ln ∣ V - 1∣ + \frac{1}{2} ln ∣ V - 3∣ = \frac{1}{2} ln \frac{V - 3}{V - 1} .$

4. $F (5) - F (4)$ evaluate karo. Kyun $F (5) - F (4)$ ? Fundamental Theorem ke anusaar, ek definite integral upper-limit value minus lower-limit value hota hai. $F (5) = \frac{1}{2} ln \frac{2}{4} = \frac{1}{2} ln \frac{1}{2}, F (4) = \frac{1}{2} ln \frac{1}{3} .$ $T = \frac{1}{2} ln \frac{1}{2} - \frac{1}{2} ln \frac{1}{3} = \frac{1}{2} ln \frac{3}{2} \approx 0.2027 s .$

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

Cell I — exam twist: inverse Laplace via partial fractions

Worked example (9)

$f (t)$ nikalo agar uska Laplace transform hai $F (s) = \frac{s + 3}{s ( s ^{2} + 2 s + 5 )}$ .

Forecast: $s$ factor ek constant deta hai, quadratic ek exponential-times-sinusoid deta hai (iske roots complex hain).

1. $s$ mein partial-fraction karo. Kyun? Standard inverse-Laplace tables sirf simple pieces jaanti hain, toh hume pehle split karna hoga — wohi template machinery, ab $s$ variable ke saath. $\frac{s + 3}{s ( s ^{2} + 2 s + 5 )} = \frac{A}{s} + \frac{B s + C}{s ^{2} + 2 s + 5} .$

2. $A$ ke liye cover-up. Kyun yahan cover-up: $s = 0$ denominator ka real root hai, toh $s$ se multiply karke $s = 0$ set karne se quadratic term mar jaata hai (usme $s \to 0$ factor hai) aur $A$ isolate ho jaata hai. $A = \frac{0 + 3}{0 + 0 + 5} = \frac{3}{5} .$

3. $B, C$ ke liye match karo. Kyun matching: quadratic ka koi real root nahi plug in karne ke liye, toh hum coefficients compare karte hain. Denominators clear karo: $s + 3 = A (s^{2} + 2 s + 5) + (B s + C) s .$

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

4. Quadratic mein complete the square karo: $s^{2} + 2 s + 5 = (s + 1)^{2} + 2^{2}$ . Kyun? Inverse-Laplace table shifted form $(s + a)^{2} + ω^{2}$ mein likhi hai, jo $e^{- a t}$ times frequency $ω$ ke sinusoid mein decode hoti hai — toh hume pehle top ko $(s + 1)$ ke around rewrite karna hoga: $- \frac{3}{5} s - \frac{1}{5} = - \frac{3}{5} (s + 1) + \frac{2}{5} .$ Toh quadratic block hai $- \frac{3}{5} \cdot \frac{s + 1}{( s + 1 ) ^{2} + 4} + \frac{2}{5} \cdot \frac{1}{( s + 1 ) ^{2} + 4} .$

5. Inverses read off karo. Kyun ye? Table: $\frac{s + 1}{( s + 1 ) ^{2} + 4} \to e^{- t} cos 2 t$ , aur $\frac{2}{( s + 1 ) ^{2} + 4} \to e^{- t} sin 2 t$ . $f (t) = \frac{3}{5} - \frac{3}{5} e^{- t} cos 2 t + \frac{2}{5} \cdot \frac{1}{2} e^{- t} sin 2 t = \frac{3}{5} - \frac{3}{5} e^{- t} cos 2 t + \frac{1}{5} e^{- t} sin 2 t .$

Verify: $t = 0$ par, $f (0) = \frac{3}{5} - \frac{3}{5} + 0 = 0$ , jo initial-value theorem $lim_{s \to \infty} s F (s) = lim \frac{s ( s + 3 )}{s ( s ^{2} + 2 s + 5 )} = 0$ se match karta hai ✓ (VERIFY).

Cell J — mixed: linear aur irreducible quadratic saath mein

Worked example (10)

$\int \frac{2 x ^{2} - x + 4}{( x - 1 ) ( x ^{2} + 4 )} d x$ Sabse general case: ek linear factor aur ek irreducible quadratic, toh dono templates ek saath aate hain aur interact karte hain.

Forecast: ek $ln ∣ x - 1∣$ (linear se), ek $ln (x^{2} + 4)$ (quadratic ka log part), aur ek $arctan$ (quadratic ka arctan part) expect karo.

1. Confirm karo ki quadratic irreducible hai aur joint template likho. Kyun? $x^{2} + 4$ ka discriminant $- 16 < 0$ hai, toh ye pura rehta hai aur ek linear top $B x + C$ maangta hai; linear factor ko ek akela constant $A$ chahiye. $\frac{2 x ^{2} - x + 4}{( x - 1 ) ( x ^{2} + 4 )} = \frac{A}{x - 1} + \frac{B x + C}{x ^{2} + 4} .$ Counting check: total unknowns $= 3 = de g Q$ ✓.

2. $A$ ke liye cover-up. Kyun linear piece ke liye cover-up kaam karta hai: $x = 1$ real root hai; $(x - 1)$ se multiply karke $x = 1$ set karne se quadratic term mar jaata hai. $A = \frac{2 ( 1 ) ^{2} - 1 + 4}{1 ^{2} + 4} = \frac{5}{5} = 1.$

3. $B, C$ ke liye match karo. Kyun matching, cover-up nahi: quadratic ka koi real root nahi plug in karne ke liye, toh hum coefficients compare karte hain. $A = 1$ ke saath denominators clear karo: $2 x^{2} - x + 4 = (x^{2} + 4) + (B x + C) (x - 1) .$ RHS expand karo $= x^{2} + 4 + B x^{2} - B x + C x - C = (1 + B) x^{2} + (C - B) x + (4 - C)$ . Match karo:

$x^{2} : 1 + B = 2 \Rightarrow B = 1$
$x^{0} : 4 - C = 4 \Rightarrow C = 0$
(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. Har piece integrate karo. Kyun ye forms? Linear ek log deta hai; $\frac{x}{x ^{2} + 4}$ ke upar (a multiple of) denominator ka derivative $2 x$ hai, toh ye substitution $u = x^{2} + 4$ se ek log hai — yahan koi arctan zaroorat nahi kyunki $C = 0$ ne koi bare constant nahi chhoda. $\int = ln ∣ x - 1∣ + \frac{1}{2} ln x^{2} + 4 + C .$ (Agar $C$ nonzero hota, toh ek term $\frac{C}{x ^{2} + 4}$ ek $\frac{C}{2} arctan \frac{x}{2}$ add karta — poora mixed answer tab teeno pieces carry karta.)

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

Recall

Recall Which cell is which?

Irreducible quadratic ke upar pure constant kya deta hai? ::: Ek pure $arctan$ (Cell C). Irreducible quadratic ke upar $x$ wala numerator kya deta hai? ::: log plus arctan (Cell D). Agar $de g P \geq de g Q$ ho toh pehli cheez kya karni chahiye? ::: Polynomial long division (Cell E). Positive discriminant wale quadratic ko kaise treat karna chahiye? ::: Do linear factors ke roop mein (Cell G). Repeated quadratic $(x^{2} + 1)^{2}$ ka template kya hai? ::: $\frac{B _{1} x + C _{1}}{x ^{2} + 1} + \frac{B _{2} x + C _{2}}{( x ^{2} + 1 ) ^{2}}$ (Cell F). Ek linear aur ek irreducible-quadratic factor wali problem ko kya chahiye? ::: dono templates $\frac{A}{x - a} + \frac{B x + C}{x ^{2} + \dots}$ ek saath (Cell J).

Ye prerequisites ek detour ke layak hain: Completing the square, Integration by substitution, Standard integrals — log and arctan, Factoring polynomials over the reals.