Partial fractions — linear, repeated, irreducible quadratic factors: Visual walkthrough

4.2.10 · D2 · Maths › Calculus II — Integration › Partial fractions — linear, repeated, irreducible quadratic

Step 0 — "Rational function" kya hoti hai, aur hum kya touch kar sakte hain?

Hum poore walkthrough mein ek hi running example trace karenge: $P (x) 3 x + 1 / Q (x) (x^{2} - x - 2) .$

$P (x) = 3 x + 1$ — top (upar wala), degree $1$ .
$Q (x) = x^{2} - x - 2$ — bottom (neeche wala), degree $2$ .

Kyunki $de g P = 1 < 2 = de g Q$ , yeh fraction proper hai — top genuinely bottom se chhota hai. Hum pehle yeh kyun check karte hain: sirf proper fractions hi purely simple pieces se bann sakti hain; zyada bada top ek leftover polynomial maangta (yeh hai Polynomial long division, alag se handle kiya gaya). Neeche har step proper maanke chalta hai.

Step 1 — Denominator ko factors mein "un-multiply" hote dekho

KYA. Hum bottom ko sabse simple possible factors ki product mein split karte hain: $x^{2} - x - 2 = (x - 2) (x + 1) .$ Yahan $x - 2$ aur $x + 1$ linear factors hain — har ek bas " $x$ minus ek number" hai. Numbers $2$ aur $- 1$ roots hain: woh $x$ ki values jo bottom ko zero banati hain (yeh hai Factoring polynomials over the reals).

KYUN. Denominator ek product hai. Do cheezein ki product sirf tab zero hoti hai jab unme se ek zero ho. Toh poori fraction bilkul do jagah "blast" hoti hai ("infinity tak ud jaati hai"): $x = 2$ aur $x = - 1$ par. Yeh do blast points woh fingerprints hain jinhe hum dhundenge.

PICTURE. Neeche, blue curve hai $y = \frac{3 x + 1}{( x - 2 ) ( x + 1 )}$ . Notice karo do vertical dashed walls — ek yellow $x = - 1$ par, ek red $x = 2$ par — jahan curve infinity tak shoot karta hai. Har wall ek factor ke zero hone se banti hai.

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

Step 2 — Shape guess karo: har explosion ke liye ek simple fraction

KYA. Hum claim karte hain ki fraction ek sum hai, har factor ke liye ek term: $\frac{3 x + 1}{( x - 2 ) ( x + 1 )} = \frac{A}{x - 2} + \frac{B}{x + 1} .$

$\frac{A}{x - 2}$ — ek piece jo sirf $x = 2$ par blast hoti hai (apni wall par).
$\frac{B}{x + 1}$ — ek piece jo sirf $x = - 1$ par blast hoti hai.
$A$ aur $B$ — do unknown numbers ("har blast kitni strong hai") jo hume abhi bhi find karne hain.

WHY exactly yeh shape. Fractions add karna reversible hai. Agar kisi ne $\frac{A}{x - 2} + \frac{B}{x + 1}$ ko common denominator par add kiya hota, toh woh wapas bottom par $(x - 2) (x + 1)$ par land karte. Toh peeche jaate hue, hamari fraction zaroor unhi denominators wale pieces se aayi hogi. Har piece ka top sirf ek constant hai kyunki har denominator degree $1$ ka hai, aur ek proper piece ko degree $0$ ka top chahiye.

PICTURE. Blue curve (poora function) do colored simple curves ka sum hai: red wala $x = 2$ ki wall ka malik hai, yellow wala $x = - 1$ ki wall ka. Walls se door, woh aaram se add ho jaate hain aur blue banaate hain.

Recall

Har simple piece ka top sirf ek constant kyun hai (na ki $A x + \dots$ )? ::: Har denominator ki degree $1$ hai, toh ek proper piece ko degree $0$ ka numerator chahiye — ek akela constant.

Step 3 — Cover-up trick, term by term dekhi gayi

KYA. $A$ find karne ke liye, dono sides ko factor $(x - 2)$ se multiply karo: $\frac{3 x + 1}{x + 1} = A + (x - 2) \cdot \frac{B}{x + 1} .$

Har piece dekho:

Left: $(x - 2)$ ne bottom wala $(x - 2)$ cancel kar diya, bacha $\frac{3 x + 1}{x + 1}$ .
Right, term 1: $\frac{A}{x - 2} \times (x - 2) = A$ — akela, koi $x$ nahi bacha.
Right, term 2: $\frac{B}{x + 1} \times (x - 2) = (x - 2) \frac{B}{x + 1}$ — isme abhi bhi ek live $(x - 2)$ hai.

KYUN. Humne situation ko engineer kiya taaki $x = 2$ par annoying $B$ -term $(x - 2) = 0$ se multiply ho aur gayab ho jaaye. Isse $A$ akela reh jaata hai: $A = \frac{3 x + 1}{x + 1}_{x = 2} = \frac{3 ( 2 ) + 1}{2 + 1} = \frac{7}{3} .$

PICTURE. Animation-frame mein dikhta hai ki $B$ -term ka contribution zero ho jaata hai jaise $x$ slide karke $2$ tak jaata hai (green curve axis tak pinch hoti hai), jabki bacha hua quantity $\frac{3 x + 1}{x + 1}$ (blue) yellow dot par clearly read hoti hai.

$(x + 1)$ cover karke $x = - 1$ set karo: $B = \frac{3 x + 1}{x - 2}_{x = - 1} = \frac{- 3 + 1}{- 1 - 2} = \frac{- 2}{- 3} = \frac{2}{3} .$

Step 4 — Split assemble karo aur sanity-check karo

KYA. $A = \frac{7}{3}$ , $B = \frac{2}{3}$ ke saath: $\frac{3 x + 1}{( x - 2 ) ( x + 1 )} = \frac{7/3}{x - 2} + \frac{2/3}{x + 1} .$

KYUN check karte hain. Do unknowns, do conditions — lekin ek safe point $x = 0$ test karke verify karte hain (dono walls se door):

Left: $\frac{1}{( - 2 ) ( 1 )} = - \frac{1}{2}$ .
Right: $\frac{7/3}{- 2} + \frac{2/3}{1} = - \frac{7}{6} + \frac{4}{6} = - \frac{3}{6} = - \frac{1}{2}$ . ✓

PICTURE. Blue (original) aur colored sum ek hi axes par draw hain; woh dono walls ko chhodkar har jagah exactly ek dusre ke upar hain. $x = 0$ par green vertical tie-line dikhati hai ki dono $- \frac{1}{2}$ de rahe hain.

Step 5 — Integrate karo: har simple wall kyun ek logarithm banti hai

KYA. $\int \frac{3 x + 1}{( x - 2 ) ( x + 1 )} d x = \frac{7}{3} ln ∣ x - 2∣ + \frac{2}{3} ln ∣ x + 1∣ + C .$

LOG KYUN? Woh akaali function jiska derivative $\frac{1}{x - a}$ ho, woh $ln ∣ x - a ∣$ hai — yeh ek standard integral hai. Toh ek fraction jo apni wall ke paas $\frac{constant}{x - a}$ jaisa behave kare, woh zaroor ek scaled log tak integrate hogi. Yahi poora faayda hai: humne ek un-integrable mess ko do instant logs se trade kar liya.

PICTURE. Har colored $\frac{1}{x - a}$ spike (upar) apni $ln ∣ x - a ∣$ curve (neeche) ke upar hai, aur $x = a$ par vertical wall woh point ban jaati hai jahan log $- \infty$ tak dive karta hai.

Step 6 — Edge case: agar bottom linear walls mein split na ho toh?

KYA. Maano bottom mein ek irreducible quadratic ho jaise $x^{2} + 1$ — jo koi real roots nahi rakhti (iska discriminant $0^{2} - 4 \cdot 1 \cdot 1 = - 4 < 0$ hai). Toh iske liye koi real wall nahi: curve $\frac{1}{x ^{2} + 1}$ kabhi blast nahi hoti.

WHY isko ek linear top $B x + C$ chahiye. Koi real root nahi, toh hum ek single constant isolate nahi kar sakte. Quadratic denominator ki degree $2$ hai, toh iske upar ek proper piece ka numerator degree $1$ ka hona chahiye: $\frac{B x + C}{x ^{2} + 1}$ . $B x$ part ek log tak integrate hota hai (Integration by substitution ke zariye, $u = x^{2} + 1$ ); bacha hua constant $C$ ek arctan tak integrate hota hai — woh smooth "no-wall" swirl, ek aur standard integral. General $x^{2} + b x + c$ ke liye Completing the square dekho.

PICTURE. Contrast panel: left mein, $\frac{1}{x - 1}$ apni real wall ke saath (→ log); right mein, $\frac{1}{x ^{2} + 1}$ , ek smooth hump bina wall ke (→ arctan). Alag geometry ⇒ alag integral.

Ek-picture summary

Ek messy blue curve, do explosion-walls → cover-up se do colored simple fractions mili → integrate karne par do logs. Yahi poori machine hai.

Recall Feynman retelling — poora walkthrough seedhe shabdon mein

Mere paas ek lumpy fraction hai. Pehle main uska bottom dekh ke use factor karta hoon — har factor ek aisi jagah mark karta hai jahan graph infinity tak shoot karta hai, ek "wall". Main guess karta hoon ki meri lumpy fraction asal mein ek sum hai, har wall ke liye ek baby fraction, aur har baby sirf apni wall par blast hoti hai. Kisi baby ki strength find karne ke liye, main cover-up trick use karta hoon: main us wall ke factor se through multiply karta hoon taaki har doosri baby ko ek aisi cheez se multiply kar doon jo bilkul us wall par zero ho jaaye — woh sab gayab ho jaate hain, aur bacha hua mujhe apna number de deta hai. Main yeh har wall ke liye karta hoon, babies ko wapas saath jodhta hoon, aur ek aasaan point test karta hoon ki woh original mein add hote hain ya nahi. Ant mein main integrate karta hoon: har $\frac{1}{x - a}$ wall bas $ln ∣ x - a ∣$ hai. Agar bottom mein ek aisa piece ho jisme koi real wall na ho (ek irreducible quadratic), toh us piece ka top linear $B x + C$ rehta hai aur yeh ek log plus ek smooth arctan swirl mein integrate hota hai. Walls → logs, no-walls → arctans. Ho gaya.

Active recall

Denominator ka ek factor graph par ek "wall" se kaise correspond karta hai?

Fraction wahan infinity tak blast hoti hai jahan bottom zero hoti hai; har factor exactly ek aisa zero deta hai.

Cover-up step mein doosra term kyun gayab ho jaata hai?

Tum

(x - a)

se multiply karte ho; doosre term mein ek live factor

(x - a)

rehta hai jo

x = a

par

0

ban jaata hai.

Hamare example mein

A

aur

B

kya hain?

A = \frac{7}{3}

over

(x - 2)

aur

B = \frac{2}{3}

over

(x + 1)

Har simple piece log tak integrate kyun hoti hai?

\int \frac{1}{x - a} d x = ln ∣ x - a ∣

woh standard integral hai jo ek linear factor ke upar lone constant se match karta hai.

Irreducible quadratic

B x + C

top kyun rakhta hai aur arctan kyun deta hai?

Iska koi real root nahi hai jise cover up karein; degree-2 bottom ko degree-1 top chahiye, aur bacha hua constant arctan tak integrate hota hai (koi real wall nahi).