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# Integration by Partial Fractions

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#### Integration by Partial Fractions - Lesson Summary

While evaluating integrals, where the integrand is a fraction of polynomials.

Integral contains P(x)/Q(x)

First, check whether the function is a proper fraction.

If the degree of the numerator is less than that of the denominator, then it is a proper fraction.

i .e Degree P(x) < Degree Q(x)

If the function is a proper fraction, then check whether the denominator can be factorised into linear or quadratic factors.

If the denominator cannot be split, then other integration methods are chosen.

If the denominator can be split, then split the function into partial fractions.

We will shortly discuss the ways to split a function into partial fractions.

Improper fraction P(x)/Q(x) Degree P(x) > Degree Q(x)

If the function is a not a proper fraction, that is, the degree of the numerator is greater than or equal to that of the denominator, then it is an improper fraction.

Thus, the degree of the numerator is greater than the degree of the denominator, and hence, it is an improper fraction.

In such cases, we convert the function into a proper fraction by dividing the numerator by the denominator.

After this, we check whether the denominator of this proper fraction can be factorised into linear or quadratic factors.

If the denominator can be split, then split P1(x)/Q(x) into partial fractions.

P(x)/Q(x) = T(x) + P1(x)/Q(x)

This is how we convert an improper fraction into a proper fraction.

Now we split a proper fraction into partial fractions.

Now  px+q / (x-a)(x-b) = A/(x-a) + B/(x-b)

px+q / (x-a)(x-b) = A(x-b)+B(x-a) /(x-a)(x-b)

px+q = A(x-b)+B(x-a)

px+q = (A+B)x - (Ab + Ba)

Comparing the coefficients of x:

p = A + B .....(1)

q = -(Ab + Ba).....(2)

(1) x b + (2)

(1) x b    bp = Ab + Bb .........(1)

(1) x b + (2): bp + q = (b - a)A

â‡’ B = (bp + q)/(b-a)

(1) x a + (2)

(1) x a: ap = Aa + Ba

(1) x a + (2): ap + q = (a - b)A

A = (ap+q)/(a-b)

Ex:

âˆ« (7x - 4)/((x-1)2(x+2)) dx

Degree of numerator: 1 < Degree of denominator: 2

(7x - 4)/((x-1)2(x+2)) = A/(x-1) + B/(x-1)2 + C/(x+2)

â‡’ (7x - 4)/((x-1)2(x+2)) = (A(x-1)(x+2) + B(x+2) + c(x-1)2)/((x-1)2(x+2))

7x - 4 = A(x-1)(x+2) + B(x+2) + c(x-1)2

â‡’ 7x - 4 = A(x2 + x -2) + B(x+2) + C(x2 - 2x +1)

â‡’ 7x - 4 = x2(A+c) + x(A+B-2C) + (-2A+2B+C)

Comparing the coefficients of x2:

A+C = 0 â€¦â€¦..(1)

Comparing the coefficients of x:

A+ B -2C = 7â€¦â€¦..(2)

Comparing the constants:

-2A + 2B + C = -4â€¦â€¦â€¦(3)

A = 2, B = 1 and C = -2

âˆ« (7x - 4)/((x-1)2(x+2)) dx = âˆ« [A/(x-1) + B/(x-1)2 + C/(x+2)] dx

= âˆ« [2/(x-1) + 1/(x-1)2 - 2/(x+2)] dx

= 2 log|x-1| - 1/(x-1) - 2log|x+2| + C ('.' âˆ«1/x2 dx = -1/x + C)

= 2 (log|x-1| - log|x+2|) - 1/(x-1) + C

= 2(log|(x-1)/(x+2)|) - 1/(x-1) + C