बीजगणित में, **आंशिक भिन्न प्रसार** (partial fraction expansion) एक विधि है जो किसी परिमेय भिन्न के अंश या हर के डेग्री (degree) को कम करने के काम आती है।

सांकेतिक रूप में, निम्नलिखित परिमेय भिन्न को आंशिक भिन्नों में तोड़ा जा सकता है-

{\displaystyle {\frac {f(x)}{g(x)}}}
जहाँ **ƒ** और **g** बहुपद (polynomials) है। इसके आंशिक भिन्न निम्नवत होंगे-

{\displaystyle \sum \_{j}{\frac {f\_{j}(x)}{g\_{j}(x)}}}
जहाँ *g**j* (*x*) बहुपद हैं और ये *g*(*x*) के गुणखण्ड हैं। 

उदाहरण -
{\displaystyle {25 \over (x+2)(x^{2}+1)^{2}}}को आंशिक भिन्नों में बदलकर निम्नलिखित प्रकार से भी लिखा जा सकता है-
{\displaystyle {\frac {1}{x+2}}+{\frac {-x+2}{x^{2}+1}}+{\frac {-5x+10}{(x^{2}+1)^{2}}}\ .}

विधि
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माना दिया हुआ भिन्न {\displaystyle R(s)={\frac {P(s)}{Q(s)}}} है तो:

विधि 1
जब दिये हुए भिन्न के हर को {\displaystyle x-a} जैसे रैखिक गुणनखण्ड हो सकें ; जहाँ n >=1

{\displaystyle R(s)={\frac {P(s)}{(x-a)^{n}}}={\frac {A1}{(x-a)}}+{\frac {A2}{(x-a)^{2}}}+...+{\frac {An}{(x-a)^{n}}}}

विधि 2
जब दिये हुए भिन्न के हर का रैखिक गुणनखण्ड न हो बल्कि {\displaystyle (x-a)^{2}+b^{2}} जैसे द्विघात गुणखण्ड हो (जहाँ n >= 1) :

{\displaystyle R(s)={\frac {P(s)}{[(x-a)^{2}+b^{2}]^{n}}}={\frac {A1}{[(x-a)^{2}+b^{2}]}}+{\frac {A2}{[(x-a)^{2}+b^{2}]^{2}}}+...+{\frac {An}{[(x-a)^{2}+b^{2}]^{n}}}}

उदाहरण
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### उदाहरण १

{\displaystyle f(x)={\frac {1}{x^{2}+2x-3}}}
Here, the denominator splits into two distinct linear factors:

{\displaystyle q(x)=x^{2}+2x-3=(x+3)(x-1)}
so we have the partial fraction decomposition

{\displaystyle f(x)={\frac {1}{x^{2}+2x-3}}={\frac {A}{x+3}}+{\frac {B}{x-1}}}
Multiplying through by *x*2 + 2*x* − 3, we have the polynomial identity

{\displaystyle 1=A(x-1)+B(x+3)}
Substituting *x* = −3 into this equation gives *A* = −1/4, and substituting *x* = 1 gives *B* = 1/4, so that

{\displaystyle f(x)={\frac {1}{x^{2}+2x-3}}={\frac {1}{4}}\left({\frac {-1}{x+3}}+{\frac {1}{x-1}}\right)}

### उदाहरण २

{\displaystyle f(x)={\frac {x^{3}+16}{x^{3}-4x^{2}+8x}}}
After long-division, we have

{\displaystyle f(x)=1+{\frac {4x^{2}-8x+16}{x^{3}-4x^{2}+8x}}=1+{\frac {4x^{2}-8x+16}{x(x^{2}-4x+8)}}}
Since (−4)2 − 4×8 = −16 < 0, the factor *x*2 − 4*x* + 8 is irreducible, and the partial fraction decomposition over the reals has the shape

{\displaystyle {\frac {4x^{2}-8x+16}{x(x^{2}-4x+8)}}={\frac {A}{x}}+{\frac {Bx+C}{x^{2}-4x+8}}}
Multiplying through by *x*3 − 4*x*2 + 8*x*, we have the polynomial identity

{\displaystyle 4x^{2}-8x+16=A(x^{2}-4x+8)+(Bx+C)x}
Taking *x* = 0, we see that 16 = 8*A*, so *A* = 2. Comparing the *x*2 coefficients, we see that 4 = *A* + *B* = 2 + *B*, so *B* = 2. Comparing linear coefficients, we see that −8 = −4*A* + *C* = −8 + *C*, so *C* = 0. Altogether,

{\displaystyle f(x)=1+2\left({\frac {1}{x}}+{\frac {x}{x^{2}-4x+8}}\right)}
The following example illustrates almost all the "tricks" one would need to use short of consulting a computer algebra system**.**

### उदाहरण ३

{\displaystyle f(x)={\frac {x^{9}-2x^{6}+2x^{5}-7x^{4}+13x^{3}-11x^{2}+12x-4}{x^{7}-3x^{6}+5x^{5}-7x^{4}+7x^{3}-5x^{2}+3x-1}}}
After long-division and factoring the denominator, we have

{\displaystyle f(x)=x^{2}+3x+4+{\frac {2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x}{(x-1)^{3}(x^{2}+1)^{2}}}}
The partial fraction decomposition takes the form

{\displaystyle {\frac {2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x}{(x-1)^{3}(x^{2}+1)^{2}}}={\frac {A}{x-1}}+{\frac {B}{(x-1)^{2}}}+{\frac {C}{(x-1)^{3}}}+{\frac {Dx+E}{x^{2}+1}}+{\frac {Fx+G}{(x^{2}+1)^{2}}}}
Multiplying through by (*x* − 1)3(*x*2 + 1)2 we have the polynomial identity

{\displaystyle {\begin{aligned}&{}\quad 2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\&=A(x-1)^{2}(x^{2}+1)^{2}+B(x-1)(x^{2}+1)^{2}+C(x^{2}+1)^{2}+(Dx+E)(x-1)^{3}(x^{2}+1)+(Fx+G)(x-1)^{3}\end{aligned}}}
Taking *x* = 1 gives 4 = 4*C*, so *C* = 1. Similarly, taking *x* = *i* gives 2 + 2*i* = (*Fi* + *G*)(2 + 2*i*), so *Fi* + *G* = 1, so *F* = 0 and *G* = 1 by equating real and imaginary parts. With *C* = *G* = 1 and *F* = 0, taking *x* = 0 we get *A* − *B* + 1 − *E* − 1 = 0, thus *E* = *A* − *B*.

We now have the identity

{\displaystyle {\begin{aligned}&{}2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\&=A(x-1)^{2}(x^{2}+1)^{2}+B(x-1)(x^{2}+1)^{2}+(x^{2}+1)^{2}+(Dx+(A-B))(x-1)^{3}(x^{2}+1)+(x-1)^{3}\\&=A((x-1)^{2}(x^{2}+1)^{2}+(x-1)^{3}(x^{2}+1))+B((x-1)(x^{2}+1)-(x-1)^{3}(x^{2}+1))+(x^{2}+1)^{2}+Dx(x-1)^{3}(x^{2}+1)+(x-1)^{3}\end{aligned}}}
Expanding and sorting by exponents of x we get

{\displaystyle {\begin{aligned}&{}2x^{6}-4x^{5}+5x^{4}-3x^{3}+x^{2}+3x\\&=(A+D)x^{6}+(-A-3D)x^{5}+(2B+4D+1)x^{4}+(-2B-4D+1)x^{3}+(-A+2B+3D-1)x^{2}+(A-2B-D+3)x\end{aligned}}}
We can now compare the coefficients and see that

{\displaystyle {\begin{aligned}A+D&=&2\\-A-3D&=&-4\\2B+4D+1&=&5\\-2B-4D+1&=&-3\\-A+2B+3D-1&=&1\\A-2B-D+3&=&3,\end{aligned}}}
with *A* = 2 − *D* and −*A* −3 *D* =−4 we get *A* = *D* = 1 and so *B* = 0, furthermore is *C* = 1, *E* = *A* − *B* = 1, *F* = 0 and *G* = 1.

The partial fraction decomposition of *ƒ*(*x*) is thus

{\displaystyle f(x)=x^{2}+3x+4+{\frac {1}{(x-1)}}+{\frac {1}{(x-1)^{3}}}+{\frac {x+1}{x^{2}+1}}+{\frac {1}{(x^{2}+1)^{2}}}.}
Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at *x=1* and at *x=i* in the above polynomial identity. (To this end, recall that the derivative at *x=a* of *(x−a)mp(x)* vanishes if *m > 1* and it is just *p(a)* if *m=1*.)
Thus, for instance the first derivative at *x=1* gives 

{\displaystyle 2\cdot 6-4\cdot 5+5\cdot 4-3\cdot 3+2+3=A\cdot (0+0)+B\cdot (4+0)+8+D\cdot 0}
that is *8 = 4B + 8* so *B=0*.

### उदाहरण ४ (residue method)

{\displaystyle f(z)={\frac {z^{2}-5}{(z^{2}-1)(z^{2}+1)}}={\frac {z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}}}
Thus, *f*(*z*) can be decomposed into rational functions whose denominators are *z*+1, *z*−1, *z*+i, *z*−i. Since each term is of power one, −1, 1, −*i* and *i* are simple poles.

Hence, the residues associated with each pole, given by

{\displaystyle {\frac {P(z\_{i})}{Q'(z\_{i})}}={\frac {z\_{i}^{2}-5}{4z\_{i}^{3}}}},
are 

{\displaystyle 1,-1,{\tfrac {3i}{2}},-{\tfrac {3i}{2}}},
respectively, and

{\displaystyle f(z)={\frac {1}{z+1}}-{\frac {1}{z-1}}+{\frac {3i}{2}}{\frac {1}{z+i}}-{\frac {3i}{2}}{\frac {1}{z-i}}}.
### उदाहरण ५ (limit method)

Limits can be used to find a partial fraction decomposition.

{\displaystyle f(x)={\frac {1}{x^{3}-1}}}
First, factor the denominator:

{\displaystyle f(x)={\frac {1}{(x-1)(x^{2}+x+1)}}}
The decomposition takes the form of

{\displaystyle {\frac {1}{(x-1)(x^{2}+x+1)}}={\frac {A}{x-1}}+{\frac {Bx+C}{x^{2}+x+1}}}
As {\displaystyle x\to 1}, the *A* term dominates, so the right-hand side approaches {\displaystyle {\frac {A}{x-1}}}. Thus, we have

{\displaystyle {\frac {1}{(x-1)(x^{2}+x+1)}}={\frac {A}{x-1}}}
{\displaystyle A=\lim \_{x\to 1}{\frac {1}{x^{2}+x+1}}={\frac {1}{3}}}
As {\displaystyle x\to \infty }, the right-hand side is

{\displaystyle \lim \_{x\to \infty }{{\frac {A}{x-1}}+{\frac {Bx+C}{x^{2}+x+1}}}={\frac {A}{x}}+{\frac {Bx}{x^{2}}}={\frac {A+B}{x}}.}
{\displaystyle {\frac {A+B}{x}}=\lim \_{x\to \infty }{\frac {1}{x^{3}-1}}=0}
Thus, {\displaystyle B=-{\frac {1}{3}}}.

At {\displaystyle x=0}, {\displaystyle -1=-A+C}. Therefore, {\displaystyle C=-{\frac {2}{3}}}.

The decomposition is thus {\displaystyle {\frac {\frac {1}{3}}{x-1}}+{\frac {-{\frac {1}{3}}x-{\frac {2}{3}}}{x^{2}+x+1}}}.

सन्दर्भ
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1. ↑  Bluman, George W. (1984). *Problem Book for First Year Calculus*. New York: Springer-Verlag. पपृ॰ 250–251.

बाहरी कड़ियाँ
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* एरिक डब्ल्यू वेइसटीन, मैथवर्ल्ड पर *Partial Fraction Decomposition*