Intuition The big pictureA complicated rational function P ( x ) Q ( x ) \dfrac{P(x)}{Q(x)} Q ( x ) P ( x ) simple pieces like A x − a \dfrac{A}{x-a} x − a A 1 x 2 + 1 \dfrac{1}{x^2+1} x 2 + 1 1 Partial fractions is the algebra of un-adding fractions: we split one ugly fraction back into the easy pieces that were combined to make it.
WHY it works: addition of fractions over a common denominator is a reversible operation. If Q ( x ) Q(x) Q ( x ) must have been built from fractions whose denominators are exactly those pieces. We just have to find the numerators.
Definition Proper rational functionP ( x ) Q ( x ) \dfrac{P(x)}{Q(x)} Q ( x ) P ( x ) proper when deg P < deg Q \deg P < \deg Q deg P < deg Q only applies to proper functions. If improper, first do polynomial long division :
P Q = ( polynomial ) + R Q , deg R < deg Q . \frac{P}{Q} = (\text{polynomial}) + \frac{R}{Q},\qquad \deg R < \deg Q. Q P = ( polynomial ) + Q R , deg R < deg Q .
HOW (the master recipe):
Make it proper (long division if needed).
Fully factor Q ( x ) Q(x) Q ( x ) linear and irreducible quadratic factors.
For each factor, write the right template of unknown numerators (rules below).
Solve for the unknowns (cover-up / matching coefficients).
Integrate each simple piece.
Intuition WHY repeated factors needall lower powers
Suppose only A ( x − a ) 2 \dfrac{A}{(x-a)^2} ( x − a ) 2 A ( x − a ) 2 (x-a)^2 ( x − a ) 2 A A A any linear A 2 + A 1 ( x − a ) A_2 + A_1(x-a) A 2 + A 1 ( x − a ) need both A 1 x − a \dfrac{A_1}{x-a} x − a A 1 A 2 ( x − a ) 2 \dfrac{A_2}{(x-a)^2} ( x − a ) 2 A 2 B x + C Bx+C B x + C
Take a simple linear factor. Write
P ( x ) ( x − a ) g ( x ) = A x − a + (rest) g ( x ) . \frac{P(x)}{(x-a)\,g(x)} = \frac{A}{x-a} + \frac{\text{(rest)}}{g(x)}. ( x − a ) g ( x ) P ( x ) = x − a A + g ( x ) (rest) . ( x − a ) (x-a) ( x − a ) P ( x ) g ( x ) = A + ( x − a ) ⋅ (rest) g ( x ) . \frac{P(x)}{g(x)} = A + (x-a)\cdot\frac{\text{(rest)}}{g(x)}. g ( x ) P ( x ) = A + ( x − a ) ⋅ g ( x ) (rest) . x = a x=a x = a ( x − a ) → 0 (x-a)\to 0 ( x − a ) → 0 A = P ( a ) g ( a ) \boxed{\,A = \frac{P(a)}{g(a)}\,} A = g ( a ) P ( a ) cover up ( x − a ) (x-a) ( x − a ) x = a x=a x = a highest-power coefficient cleanly; lower repeated/quadratic coefficients still need matching.)
Worked example (1) Distinct linear factors∫ 3 x + 1 x 2 − x − 2 d x \int \frac{3x+1}{x^2-x-2}\,dx ∫ x 2 − x − 2 3 x + 1 d x Factor: x 2 − x − 2 = ( x − 2 ) ( x + 1 ) x^2-x-2=(x-2)(x+1) x 2 − x − 2 = ( x − 2 ) ( x + 1 ) deg 1 < deg 2 \deg 1<\deg 2 deg 1 < deg 2 3 x + 1 ( x − 2 ) ( x + 1 ) = A x − 2 + B x + 1 \dfrac{3x+1}{(x-2)(x+1)}=\dfrac{A}{x-2}+\dfrac{B}{x+1} ( x − 2 ) ( x + 1 ) 3 x + 1 = x − 2 A + x + 1 B Cover-up for A A A cover ( x − 2 ) (x-2) ( x − 2 ) x = 2 x=2 x = 2 A = 3 ( 2 ) + 1 2 + 1 = 7 3 A=\dfrac{3(2)+1}{2+1}=\dfrac{7}{3} A = 2 + 1 3 ( 2 ) + 1 = 3 7 Why? the B B B ( x − 2 ) (x-2) ( x − 2 ) x = 2 x=2 x = 2 Cover-up for B B B cover ( x + 1 ) (x+1) ( x + 1 ) x = − 1 x=-1 x = − 1 B = 3 ( − 1 ) + 1 − 1 − 2 = − 2 − 3 = 2 3 B=\dfrac{3(-1)+1}{-1-2}=\dfrac{-2}{-3}=\dfrac{2}{3} B = − 1 − 2 3 ( − 1 ) + 1 = − 3 − 2 = 3 2 Integrate: ∫ = 7 3 ln ∣ x − 2 ∣ + 2 3 ln ∣ x + 1 ∣ + C . \int = \tfrac{7}{3}\ln|x-2| + \tfrac{2}{3}\ln|x+1| + C. ∫ = 3 7 ln ∣ x − 2∣ + 3 2 ln ∣ x + 1∣ + C . Why log? each A x − a \frac{A}{x-a} x − a A A ln ∣ x − a ∣ A\ln|x-a| A ln ∣ x − a ∣
Worked example (2) Repeated linear factorx + 4 ( x − 1 ) 2 ( x + 2 ) = A x − 1 + B ( x − 1 ) 2 + C x + 2 . \frac{x+4}{(x-1)^2(x+2)} = \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+2}. ( x − 1 ) 2 ( x + 2 ) x + 4 = x − 1 A + ( x − 1 ) 2 B + x + 2 C . B B B ( x − 1 ) 2 (x-1)^2 ( x − 1 ) 2 x = 1 x=1 x = 1 B = 1 + 4 1 + 2 = 5 3 B=\dfrac{1+4}{1+2}=\dfrac{5}{3} B = 1 + 2 1 + 4 = 3 5 Why this works for the top power: multiplying by ( x − 1 ) 2 (x-1)^2 ( x − 1 ) 2 B B B x = 1 x=1 x = 1 C C C ( x + 2 ) (x+2) ( x + 2 ) x = − 2 x=-2 x = − 2 C = − 2 + 4 ( − 3 ) 2 = 2 9 C=\dfrac{-2+4}{(-3)^2}=\dfrac{2}{9} C = ( − 3 ) 2 − 2 + 4 = 9 2 A A A A A A x 2 x^2 x 2 x + 4 x+4 x + 4 x 2 x^2 x 2 A + C = 0 ⇒ A = − C = − 2 9 A+C=0\Rightarrow A=-C=-\dfrac{2}{9} A + C = 0 ⇒ A = − C = − 9 2 Why pick x 2 x^2 x 2 it's the only place A A A C C C B B B ∫ = − 2 9 ln ∣ x − 1 ∣ − 5 3 ⋅ 1 x − 1 + 2 9 ln ∣ x + 2 ∣ + C . \int = -\tfrac{2}{9}\ln|x-1| - \tfrac{5}{3}\cdot\frac{1}{x-1} + \tfrac{2}{9}\ln|x+2| + C. ∫ = − 9 2 ln ∣ x − 1∣ − 3 5 ⋅ x − 1 1 + 9 2 ln ∣ x + 2∣ + C . Why no log for B B B ∫ ( x − 1 ) − 2 d x = − ( x − 1 ) − 1 \int (x-1)^{-2}dx = -(x-1)^{-1} ∫ ( x − 1 ) − 2 d x = − ( x − 1 ) − 1 ≠ − 1 \neq -1 = − 1
Worked example (3) Irreducible quadratic factor2 x 2 + x + 1 ( x + 1 ) ( x 2 + 1 ) = A x + 1 + B x + C x 2 + 1 . \frac{2x^2+x+1}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}. ( x + 1 ) ( x 2 + 1 ) 2 x 2 + x + 1 = x + 1 A + x 2 + 1 B x + C . x 2 + 1 x^2+1 x 2 + 1 0 2 − 4 ( 1 ) ( 1 ) = − 4 < 0 0^2-4(1)(1)=-4<0 0 2 − 4 ( 1 ) ( 1 ) = − 4 < 0 A A A x = − 1 x=-1 x = − 1 A = 2 ( 1 ) − 1 + 1 ( − 1 ) 2 + 1 = 2 2 = 1. A=\dfrac{2(1)-1+1}{(-1)^2+1}=\dfrac{2}{2}=1. A = ( − 1 ) 2 + 1 2 ( 1 ) − 1 + 1 = 2 2 = 1. B , C B,C B , C 2 x 2 + x + 1 = A ( x 2 + 1 ) + ( B x + C ) ( x + 1 ) . 2x^2+x+1 = A(x^2+1)+(Bx+C)(x+1). 2 x 2 + x + 1 = A ( x 2 + 1 ) + ( B x + C ) ( x + 1 ) . A = 1 A=1 A = 1 = x 2 + 1 + B x 2 + B x + C x + C = ( 1 + B ) x 2 + ( B + C ) x + ( 1 + C ) . =x^2+1+Bx^2+Bx+Cx+C=(1+B)x^2+(B+C)x+(1+C). = x 2 + 1 + B x 2 + B x + C x + C = ( 1 + B ) x 2 + ( B + C ) x + ( 1 + C ) . x 2 : 1 + B = 2 ⇒ B = 1 x^2:\,1+B=2\Rightarrow B=1 x 2 : 1 + B = 2 ⇒ B = 1 1 + C = 1 ⇒ C = 0 1+C=1\Rightarrow C=0 1 + C = 1 ⇒ C = 0 x x x B + C = 1 B+C=1 B + C = 1 2 x 2 + x + 1 ( x + 1 ) ( x 2 + 1 ) = 1 x + 1 + x x 2 + 1 . \frac{2x^2+x+1}{(x+1)(x^2+1)}=\frac{1}{x+1}+\frac{x}{x^2+1}. ( x + 1 ) ( x 2 + 1 ) 2 x 2 + x + 1 = x + 1 1 + x 2 + 1 x . Integrate: ∫ = ln ∣ x + 1 ∣ + 1 2 ln ( x 2 + 1 ) + C . \int=\ln|x+1|+\tfrac12\ln(x^2+1)+C. ∫ = ln ∣ x + 1∣ + 2 1 ln ( x 2 + 1 ) + C . Why? ∫ x x 2 + 1 d x = 1 2 ln ( x 2 + 1 ) \int\frac{x}{x^2+1}dx=\tfrac12\ln(x^2+1) ∫ x 2 + 1 x d x = 2 1 ln ( x 2 + 1 ) u = x 2 + 1 u=x^2+1 u = x 2 + 1 C C C x 2 + 1 x^2+1 x 2 + 1 arctan x \arctan x arctan x
For B x + C x 2 + b x + c \dfrac{Bx+C}{x^2+bx+c} x 2 + b x + c B x + C split into two ideas:
The log part: engineer the numerator to be the derivative of the denominator ( 2 x + b ) (2x+b) ( 2 x + b ) B 2 ln ( x 2 + b x + c ) \tfrac{B}{2}\ln(x^2+bx+c) 2 B ln ( x 2 + b x + c ) The arctan part: the constant leftover, after completing the square x 2 + b x + c = ( x + b 2 ) 2 + d 2 x^2+bx+c=(x+\tfrac{b}{2})^2 + d^2 x 2 + b x + c = ( x + 2 b ) 2 + d 2 1 d arctan x + b / 2 d \tfrac{1}{d}\arctan\!\frac{x+b/2}{d} d 1 arctan d x + b /2
∫ x + 3 x 2 + 2 x + 5 d x \displaystyle\int\frac{x+3}{x^2+2x+5}dx ∫ x 2 + 2 x + 5 x + 3 d x 2 x + 2 2x+2 2 x + 2 x + 3 = 1 2 ( 2 x + 2 ) + 2 x+3=\tfrac12(2x+2)+2 x + 3 = 2 1 ( 2 x + 2 ) + 2 ∫ = 1 2 ln ( x 2 + 2 x + 5 ) + ∫ 2 ( x + 1 ) 2 + 4 d x = 1 2 ln ( x 2 + 2 x + 5 ) + 2 ⋅ 1 2 arctan x + 1 2 + C . \int=\tfrac12\ln(x^2+2x+5)+\int\frac{2}{(x+1)^2+4}dx=\tfrac12\ln(x^2+2x+5)+2\cdot\tfrac12\arctan\tfrac{x+1}{2}+C. ∫ = 2 1 ln ( x 2 + 2 x + 5 ) + ∫ ( x + 1 ) 2 + 4 2 d x = 2 1 ln ( x 2 + 2 x + 5 ) + 2 ⋅ 2 1 arctan 2 x + 1 + C . Why split like that? it isolates an exact-derivative log plus a pure complete-the-square arctan — no guessing.
Common mistake Forgetting to make it properTempting because the template machinery looks ready to go. Why it fails: if deg P ≥ deg Q \deg P\ge\deg Q deg P ≥ deg Q Fix: long-divide first; partial-fraction the remainder only.
A ( x − a ) 2 \dfrac{A}{(x-a)^2} ( x − a ) 2 A Feels right — there's one factor, so one fraction. Why wrong: one term only covers a constant numerator; you lose the linear freedom (see intuition box). Fix: include every power A 1 x − a + A 2 ( x − a ) 2 \dfrac{A_1}{x-a}+\dfrac{A_2}{(x-a)^2} x − a A 1 + ( x − a ) 2 A 2
Common mistake Putting justA x 2 + b x + c \dfrac{A}{x^2+bx+c} x 2 + b x + c A Feels right by analogy with linear factors. Why wrong: numerator degree must be one less than denominator; a quadratic denominator needs a linear top B x + C Bx+C B x + C A A A x x x Fix: always B x + C Bx+C B x + C
Common mistake Treating a reducible quadratic as irreduciblex 2 − 5 x + 6 x^2-5x+6 x 2 − 5 x + 6 B x + C x 2 − 5 x + 6 \frac{Bx+C}{x^2-5x+6} x 2 − 5 x + 6 B x + C Why wrong: it factors ( x − 2 ) ( x − 3 ) (x-2)(x-3) ( x − 2 ) ( x − 3 ) Fix: check the discriminant; factor whenever ≥ 0 \ge 0 ≥ 0
Recall Feynman: explain it to a 12-year-oldImagine someone mixed three simple juices into one big cup and you must figure out the recipe. Partial fractions is tasting the mix and saying "ah, this much apple, this much orange, this much grape." The big messy fraction is the mixed cup; the simple fractions A x − a \frac{A}{x-a} x − a A ln \ln ln ln \ln ln arctan \arctan arctan
Mnemonic Remember the templates"Linear→Lone constant, Repeated→Run every power down, Quadratic→Quite linear top (Bx+C)."
And for integrating quadratics: "Log the derivative, arctan the square."
When can you apply partial fractions directly? Only when proper,
deg P < deg Q \deg P<\deg Q deg P < deg Q ; otherwise long-divide first.
Total number of unknown constants in the decomposition equals? deg Q \deg Q deg Q (the degree of the denominator).
Template for repeated linear factor ( x − a ) 3 (x-a)^3 ( x − a ) 3 A 1 x − a + A 2 ( x − a ) 2 + A 3 ( x − a ) 3 \frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\frac{A_3}{(x-a)^3} x − a A 1 + ( x − a ) 2 A 2 + ( x − a ) 3 A 3 .
Template for an irreducible quadratic factor ( x 2 + b x + c ) (x^2+bx+c) ( x 2 + b x + c ) B x + C x 2 + b x + c \frac{Bx+C}{x^2+bx+c} x 2 + b x + c B x + C (linear numerator).
State the cover-up rule for P ( x − a ) g \frac{P}{(x-a)g} ( x − a ) g P A = P ( a ) g ( a ) A=\frac{P(a)}{g(a)} A = g ( a ) P ( a ) : cover
( x − a ) (x-a) ( x − a ) , set
x = a x=a x = a .
Why does cover-up work? Multiply by
( x − a ) (x-a) ( x − a ) and set
x = a x=a x = a ; every other term keeps a factor
( x − a ) → 0 (x-a)\to0 ( x − a ) → 0 .
How do you check a quadratic is irreducible? Discriminant
b 2 − 4 c < 0 b^2-4c<0 b 2 − 4 c < 0 (no real roots).
∫ A x − a d x = ? \int\frac{A}{x-a}dx=? ∫ x − a A d x = ? A ln ∣ x − a ∣ + C A\ln|x-a|+C A ln ∣ x − a ∣ + C .
∫ A ( x − a ) 2 d x = ? \int\frac{A}{(x-a)^2}dx=? ∫ ( x − a ) 2 A d x = ? − A x − a + C -\frac{A}{x-a}+C − x − a A + C (power rule, not log).
Strategy to integrate B x + C x 2 + b x + c \frac{Bx+C}{x^2+bx+c} x 2 + b x + c B x + C Make numerator
∝ ( 2 x + b ) \propto(2x+b) ∝ ( 2 x + b ) for the log part; complete the square for the arctan part.
Why must a quadratic factor get B x + C Bx+C B x + C A A A Numerator degree must be one less than denominator to span all proper fractions.
Polynomial long division — prerequisite to make functions proper.Integration by substitution — used for the 1 2 ln ( x 2 + … ) \frac{1}{2}\ln(x^2+\dots) 2 1 ln ( x 2 + … ) Completing the square — converts quadratics to arctan form.Standard integrals — log and arctan — ∫ d x x 2 + a 2 = 1 a arctan x a \int\frac{dx}{x^2+a^2}=\frac1a\arctan\frac xa ∫ x 2 + a 2 d x = a 1 arctan a x Laplace transforms — inverse via partial fractions — same algebra reused.Factoring polynomials over the reals — supplies the factor list.
hard to integrate directly
linear gives log, quadratic gives arctan
Fully factor Q over reals
Cover-up or match coefficients
Cover-up rule A=P a over g a
Intuition Hinglish mein samjhoDekho, partial fractions ka idea bilkul simple hai: jab do-teen chhoti fractions ko common denominator pe add karte ho, ek badi ugly fraction ban jaati hai. Integration mein humein ulta karna hai — us badi fraction ko wapas chhoti simple pieces mein todna, kyunki simple pieces jaldi integrate ho jaate hain. A x − a \frac{A}{x-a} x − a A ln \ln ln ln \ln ln arctan \arctan arctan
Sabse pehle check karo fraction proper hai ya nahi — matlab upar ki degree neeche se choti honi chahiye. Agar nahi, to pehle polynomial long division karo. Phir denominator ko fully factor karo: linear ( x − a ) (x-a) ( x − a ) ( x − a ) k (x-a)^k ( x − a ) k ( x 2 + b x + c ) (x^2+bx+c) ( x 2 + b x + c ) template likho: linear ke liye sirf constant A A A A 1 x − a + A 2 ( x − a ) 2 + . . . \frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+... x − a A 1 + ( x − a ) 2 A 2 + ... B x + C Bx+C B x + C
Constants nikalne ke liye cover-up trick use karo: ( x − a ) (x-a) ( x − a ) x = a x=a x = a
Quadratic piece integrate karne se mat daro. Numerator ko aise tod do ki ek part denominator ka derivative ( 2 x + b ) (2x+b) ( 2 x + b ) ln \ln ln complete the square karke arctan \arctan arctan
