Partial fractions — linear, repeated, irreducible quadratic factors: Question bank

4.2.10 · D5 · Maths › Calculus II — Integration › Partial fractions — linear, repeated, irreducible quadratic

Neeche ke chaar sections cover karte hain: kya sach hai (aur kyun), log kahan galti karte hain, har rule ke peeche ki gehri "kyun", aur strange boundary cases.

Sach ya jhooth — justify karo

\frac{x ^{3} + 1}{x ^{2} - 1}

fraction ko jaise hai waise partial-fraction kiya ja sakta hai.

Jhooth — yeh improper hai (

de g 3 \geq de g 2

), isliye koi bhi proper pieces ka sum iske barabar nahi ho sakta; pehle Polynomial long division karna hoga, phir remainder ko decompose karo.

Real polynomials ki har rational function ko linear aur quadratic pieces mein split kiya ja sakta hai.

Sach — reals ke upar har polynomial linear aur irreducible quadratic factors mein factor hota hai (Factoring polynomials over the reals), isliye templates ko sirf yahi do block types chahiye hote hain.

\frac{B x + C}{x ^{2} + 1}

integrate hoke pure arctan deta hai.

Jhooth — sirf constant part

C

arctan deta hai;

B x

part ek exact-derivative log

\frac{B}{2} ln (x^{2} + 1)

hai, isliye generally tumhe dono — ek log aur ek arctan — milte hain.

Repeated factor

(x - a)^{3}

exactly teen unknown constants contribute karta hai.

Sach — tumhe har power ke liye ek term chahiye:

\frac{A _{1}}{x - a} + \frac{A _{2}}{( x - a ) ^{2}} + \frac{A _{3}}{( x - a ) ^{3}}

, aur saare blocks mein total unknown count hamesha

de g Q

ke barabar hota hai.

\frac{1}{( x - 2 ) ^{2}}

integrate hoke logarithm deta hai.

Jhooth — power rule se

- \frac{1}{x - 2}

milta hai; log tabhi milta hai jab exponent exactly

- 1

ho, jo ek squared factor nahi hota.

Agar do proper fractions infinitely many

x

ke liye barabar hain, toh unke partial-fraction constants match karne chahiye.

Sach — decomposition unique hoti hai, isliye equal rational functions identical numerators force karte hain aur isliye identical constants; yahi wajah hai ki coefficient-matching valid hai.

x^{2} + 4 x + 4

ko ek

\frac{B x + C}{x ^{2} + 4 x + 4}

block ki zaroorat hai.

Jhooth — iska discriminant

16 - 16 = 0

hai, isliye yeh

(x + 2)^{2}

ke roop mein factor hota hai: ek repeated linear factor, jise

\frac{A _{1}}{x + 2} + \frac{A _{2}}{( x + 2 ) ^{2}}

chahiye, quadratic block nahi.

Error dhundo

Ek student likhta hai

\frac{5}{( x - 1 ) ^{2}} = \frac{A}{( x - 1 ) ^{2}}

aur kuch nahi — kya missing hai?

\frac{A _{1}}{x - 1}

term missing hai; akela constant top sirf constant numerators tak pahunch sakta hai, isliye woh linear freedom ko represent nahi kar sakta jo ek repeated factor genuinely allow karta hai (parent ka "why repeated" box dekho).

Ek student

\frac{2 x}{x ^{2} + x} = \frac{B x + C}{x ^{2} + x}

set karta hai, bottom ko irreducible quadratic maan kar.

Error —

x^{2} + x = x (x + 1)

reducible hai, isliye do linear blocks

\frac{A}{x} + \frac{B}{x + 1}

hone chahiye; "quadratic top" assume karne se pehle hamesha factorability test karo.

Ek student

\frac{P}{( x - 1 ) ^{2} g}

mein

(x - 1)

cover up karta hai aur

x = 1

set karke

\frac{A}{x - 1}

constant find karta hai.

Error — highest power par cover-up se

\frac{A _{2}}{( x - 1 ) ^{2}}

constant cleanly milta hai, lekin lower

\frac{A _{1}}{x - 1}

constant ka koi simple cover-up nahi hai aur coefficient-matching chahiye.

Koi

\frac{B x + C}{x ^{2} + 2 x + 5}

ko pehle completing the square karke integrate karta hai aur

B x

term bhool jaata hai.

Error — completing the square sirf arctan half mein help karta hai; pehle exact-derivative piece

\frac{B}{2} \cdot \frac{2 x + 2}{x ^{2} + 2 x + 5}

(ek log) alag karni hogi, phir jo bachta hai uspe completing the square karo.

Ek student

x^{2} - 5 x + 6

ko irreducible maan leta hai kyunki "yeh ek quadratic hai".

Error — discriminant

25 - 24 = 1 \geq 0

hai, isliye yeh

(x - 2) (x - 3)

mein factor hota hai; "quadratic hona" test nahi hai, koi real roots na hona test hai.

Koi improper

\frac{x ^{2}}{x ^{2} - 1} = \frac{A}{x - 1} + \frac{B}{x + 1}

directly likhta hai.

Error —

de g P = de g Q

hai isliye yeh improper hai; long division pehle

1 + \frac{1}{x ^{2} - 1}

deta hai, aur sirf remainder

\frac{1}{x ^{2} - 1}

decompose hota hai.

Why questions

Quadratic ke upar numerator

B x + C

kyun hona chahiye, sirf constant kyun nahi?

Numerator degree denominator se ek kam honi chahiye; degree-2 bottom ko degree-1 top chahiye taaki har possible fraction represent ho sake, jisme woh bhi shamil hain jinke true numerator mein

x

term hota hai.

Cover-up trick sirf ek term ko kyun rehne deti hai aur baaki sab zero ho jaate hain?

(x - a)

se multiply karne par us term ka constant bare reh jaata hai jabki har doosra term abhi bhi

(x - a)

ka factor carry karta hai, jo

x = a

substitute karne par

0

ho jaata hai — Integration by substitution-style limit reasoning.

\int \frac{A}{x - a} d x

logarithm kyun deta hai?

Kyunki

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

;

\frac{1}{x - a}

shape precisely ek log ka derivative hai (Standard integrals — log and arctan), isliye linear pieces integrate karna itna aasaan hota hai.

Completing the square ek quadratic denominator ko arctan mein kyun convert karta hai?

Completing the square se

x^{2} + b x + c

(x + \frac{b}{2})^{2} + d^{2}

likha jaata hai, jo standard form

\frac{1}{u ^{2} + d ^{2}}

se match karta hai jiska antiderivative

\frac{1}{d} arctan \frac{u}{d}

hai.

Unknown constants ki total count hamesha

de g Q

kyun hoti hai?

Har factor exactly utne constants contribute karta hai jitni uski degree hai (

(x - a)^{k}

k

milte hain, quadratic se

2

milte hain), aur factor degrees ka sum

de g Q

hota hai — yeh tumhara built-in bookkeeping check hai.

Partial fractions reversible kyun hai — hum kaise jaante hain ki pieces exist karti hain?

Common denominator par fractions jodna ek reversible algebraic operation hai; kyunki

Q

exactly un pieces mein factor hota hai, original fraction zaroor un denominators wale fractions se assemble hui hogi.

Linear factor ke liye completing the square kyun unnecessary hai?

Linear denominator already simplest shape

\frac{1}{x - a}

mein hota hai jiska integral directly ek log hai; koi square complete karne ko nahi hota kyunki koi quadratic curvature nahi hai jise unwind karna ho.

Wohi partial-fraction machinery inverse Laplace transforms ko kyun power karti hai?

Laplace transforms — inverse via partial fractions ek rational transform ko standard pieces mein split karta hai jinke inverses known table entries hain, exactly waise hi jaise hum yahan log/arctan-integrable pieces mein split karte hain.

Edge cases

Agar

de g P = de g Q

exactly ho (strictly greater nahi)?

Tab bhi improper hai — long division ek constant plus proper remainder deta hai; degrees ka barabar hona pehle se hi

de g P < de g Q

requirement violate karta hai.

Agar tumne galti se

\frac{B x + C}{x ^{2} + 1}

block include kar liya jab bottom actually

(x - i) (x + i)

-reducible over reals ho... lekin actually nahi hota. Kya

x^{2} + 1

kabhi reals ke upar reducible hota hai?

Nahi —

x^{2} + 1

ka discriminant

- 4 < 0

hai aur koi real roots nahi hain, isliye yeh reals ke upar irreducible hai aur correctly

B x + C

block leta hai; yeh sirf complex numbers par "factor" hota hai, jo hum yahan use nahi karte.

Jab numerator

0

ho toh decomposition kya hogi?

Har constant

0

aayega, isliye pura fraction

0

hai; machinery tab bhi consistently chalti hai aur trivial answer return karti hai.

Repeated quadratic

(x^{2} + b x + c)^{2}

ke liye kitne constants aate hain?

Chaar — ek

\frac{B _{1} x + C _{1}}{x ^{2} + b x + c} + \frac{B _{2} x + C _{2}}{( x ^{2} + b x + c ) ^{2}}

block,

de g Q

contribution

2 \times 2 = 4

ke saath match karta hai.

Agar ek "quadratic" denominator

(x + 2)^{2}

jaisa perfect square ho — quadratic hai ya repeated-linear?

Repeated linear — iska ek real double root hai, isliye

\frac{A _{1}}{x + 2} + \frac{A _{2}}{( x + 2 ) ^{2}}

use karo;

B x + C

template un quadratics ke liye reserved hai jinka koi real root nahi hota.

Kya cover-up kabhi irreducible quadratic block ke upar constant de sakta hai?

Nahi — cover-up ko substitute karne ke liye ek real root chahiye, aur irreducible quadratics mein koi hota nahi; woh constants

B, C

denominators clear karke aur coefficients match karke hi find hote hain.

Agar wohi linear factor numerator aur denominator dono mein ho, jaise

\frac{x - 1}{( x - 1 ) ( x + 2 )}

Pehle cancel karo taaki

\frac{1}{x + 2}

mile; cancel na karna ek spurious

\frac{A}{x - 1}

term invent karta hai jo simply

0

aa jayega lekin kaam waste karta hai aur removable point mask kar sakta hai.

Recall Ek-line self-test

Agar koi tumhe $\frac{P}{Q}$ de, toh koi bhi template likhne se pehle pehle do checks kya hain? Pehla check ::: Kya yeh proper hai? Agar nahi, toh long division karo. Doosra check ::: Kya har quadratic factor genuinely irreducible hai (discriminant $< 0$ )? Jo nahi hain unhe factor karo.