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Integration Using Partial Fractions




Rational Partial Fractions

  • Rational Function is of the form
    $f(x) = \frac {p(x)}{q(x)}, q(x) \ne 0$
  • If the degree of p(x) is less than degree of q(x), it is called the Proper rational fraction
  • If the degree of p(x) is greater than degree of q(x), it is called the Improper rational fraction
  • We can reduce Improper rational fraction to Proper fraction function as
    $f(x) = \frac {p(x)}{q(x)} = t(x) +\frac {p'(x)}{q(x)} $
  • As we know how to integrate polynomials, the integration of any rational function is reduced to the integration of a proper rational function

Integration Using Partial Fractions

we can find the integration of a proper rational function using below technique
We can decompose the partial fraction using below technique and then integration can be carried out using known formulas
A. $\frac {px +q}{(x-a)(x-b)}= \frac {A}{x-a} + \frac {B}{x-b} $
and $ \int \frac {px +q}{(x-a)(x-b)} dx =\int  \left \{  \frac {A}{x-a} + \frac {B}{x-b}   \right \} dx$

B. $\frac {px^2 +qx + r}{(x-a)(x-b)(x-c)} = \frac {A}{x-a} + \frac {B}{x-b}  + \frac {C}{x-c}$
and $ \int \frac {px^2 +qx + r}{(x-a)(x-b)(x-c)} dx =\int  \left \{  \frac {A}{x-a} + \frac {B}{x-b}  + \frac {C}{x-c}  \right \} dx$

C $\frac {px +q}{(x-a)^2} = \frac {A}{x-a} + \frac {B}{(x-a)^2}$
and $ \int \frac {px +q}{(x-a)^2} dx =\int  \left \{  \frac {A}{x-a} + \frac {B}{(x-a)^2}   \right \} dx$

D $\frac {px^2 +qx + r}{(x-a)^2(x-c)}=\frac {A}{x-a} + \frac {B}{(x-a)^2}  + \frac {C}{x-c}$
and $ \int \frac {px^2 +qx + r}{(x-a)^2(x-c)} dx =\int  \left \{  \frac {A}{x-a} + \frac {B}{(x-a)^2}  + \frac {C}{x-c}  \right \} dx$

E. $ \frac {px^2 +q+r}{(x-a)(x^2 + bx +c)}= \frac {A}{x-a} + \frac {Bx +C}{x^2 + bx +c} $
and $ \int \frac {px^2 +q+r}{(x-a)(x^2 + bx +c)} dx =\int  \left \{  \frac {A}{x-a} + \frac {Bx +C}{x^2 + bx +c}   \right \} dx$
where $x^2 + bx +c$ is a irreducible quadratic

Solved examples

Example 1
Integrate:
\[ \int \frac{1}{x^2 - 1} \, dx \]
Solution
The denominator can be factored as:
\[ x^2 - 1 = (x-1)(x+1) \]
Express the integrand as a sum of partial fractions:
\[ \frac{1}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1} \]
Expanding and comparing coefficients:
For \( x = 1 \): \( A = \frac{1}{2} \)
For \( x = -1 \): \( B = -\frac{1}{2} \)
Thus, the integral becomes:
\[ \frac{1}{2} \int \frac{dx}{x-1} - \frac{1}{2} \int \frac{dx}{x+1} \]
\[ = \frac{1}{2} \ln |x-1| - \frac{1}{2} \ln |x+1| + C \]

Example 2
Integrate:
\[ \int \frac{2x+1}{x^2 + x - 2} \, dx \]
Solution
The denominator can be factored as: \[ x^2 + x - 2 = (x+2)(x-1) \]
Express the integrand as a sum of partial fractions:
\[ \frac{2x+1}{x^2 + x - 2} = \frac{A}{x+2} + \frac{B}{x-1} \]
Expanding and comparing coefficients:
For \( x = -2 \): \( A = \frac{3}{3} = 1 \)
For \( x = 1 \): \( B = 1 \)

Thus, the integral becomes:
\[ \int \frac{dx}{x+2} + \int \frac{dx}{x-1} \]
\[ = \ln |x+2| + \ln |x-1| + C \]

Example 3
\[ \int \frac{3x^2 - 2x + 4}{x^3 - x^2 - 6x} \, dx \]
Solution
The denominator can be factored as:
\[ x^3 - x^2 - 6x = x(x^2 - x - 6) = x(x-3)(x+2) \]
Express the integrand as a sum of partial fractions:
\[ \frac{3x^2 - 2x + 4}{x^3 - x^2 - 6x} = \frac{A}{x} + \frac{B}{x-3} + \frac{C}{x+2} \] Multiplying through by the denominator and equating coefficients will yield values for A, B, and C. Upon solving, you can express the integrand as the sum of simpler fractions and integrate each one separately.


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