Class 12 Maths Integrals Integration by Partial Fractions

Integration by Partial Fractions

We know that a rational function is the ratio of two polynomials in the form P(x)/Q(x), where P (x) and Q(x) are polynomials in x and Q(x) ≠ 0. If degree of P(x) is less than degree of Q(x), then it is called proper rational function otherwise it is improper. We can convert an improper rational function into proper rational function by division method.

So, if P(x)/Q(x) is improper, then P(x)/Q(x) = T(x) + P1(x)/Q(x)

Now to calculate ʃ [P(x)/Q(x)] dx where P(x)/Q(x) is a proper rational function, we first convert it into partial faction suing the method as shown in the given figure and then integrate using standard formula.

Problem: Find the integral

x/{(x - 1)(x - 2)(x - 3)}

Solution:

Let x/{(x - 1)(x - 2)(x - 3)} = A/(x - 1) + B/(x - 2) + C/(x - 3)

=> x = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)

Equating the coefficients of x2, x and constant term, we get

A + B + C = 0

-5A – 4B – 3C = 1

6A + 3B + 2C = 0

Solving these equations, we get

A = 1/2, B = −2, and C = 3/2

So, x/{(x - 1)(x - 2)(x - 3)} = 1/2(x - 1) – 2/(x - 2) + 3/2(x - 3)

Now, ʃ [x/{(x - 1)(x - 2)(x - 3)}]dx = ʃ dx/2(x - 1) – ʃ dx/2(x - 2) + ʃ 3dx/2(x - 3)

=> ʃ [x/{(x - 1)(x - 2)(x - 3)}]dx = (1/2) * ʃ dx/(x - 1) – 2 ʃ dx/(x - 2) + (3/2) * ʃ dx/(x - 3)

=> ʃ [(3x - 1)/{(x - 1)(x - 2)(x - 3)}]dx = (1/2) * log|x – 1| – 2 log|x – 2| + (3/2) * log|x – 3| + C

Problem: Find the integral

x/{(x2 + 1)(x - 1)}

Solution:

Let x/{(x2 + 1)(x - 1)} = (Ax + B)/(x2 + 1) + C/(x - 1)

=> x = (Ax + B)(x - 1) + C(x2 + 1)

=> x = Ax2 – Ax + Bx – B + Cx2 + C

Equating the coefficients of x2, x and constant terms, we get

A + C = 0

-A + B = 1

-B + C = 0

On solving these equations, we get

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

From equation 1, we get

x/{(x2 + 1)(x - 1)} = (-x/2 + 1/2)/(x2 + 1) + 1/2(x - 1)

x/{(x2 + 1)(x - 1)} = (-x/2)/(x2 + 1) + (1/2)/(x2 + 1) + 1/2(x - 1)

Now, ʃ [x/{(x2 + 1)(x - 1)} dx = (-1/2)ʃ [x/(x2 + 1)] + (1/2)ʃ [1/(x2 + 1)] + (1/2)ʃ dx/(x - 1)

= (-1/4)ʃ [2x/(x2 + 1)] dx + (1/2) * tan-1 x + (1/2) * log|x - 1| + C

Let x2 + 1 = t

=> 2x dx = dt

So, ʃ [2x/(x2 + 1)] = ʃ dt/t

= log|t|

= log|x2 + 1|

Now, ʃ [x/{(x2 + 1)(x - 1)}dx = (-1/4) log|x2 + 1| + (1/2)tan-1 x + (1/2)log|x - 1| + C

= (1/2)log|x - 1|- (1/4) log|x2 + 1| + (1/2)tan-1 x + C

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