Integration by Partial Fractions

Integration by Partial Fractions is one of the methods of integration, which is used to find the integral of rational functions. In Partial Fraction decomposition, an improper-looking rational function is decomposed into the sum of various proper rational functions.

If f(x) and g(x) are polynomials such that g(x) ≠ 0, then f(x)/g(x) is called a Rational function.

• If degree f(x) < degree g(x), then f(x)/g(x) is called a proper rational function.
• If degree f(x) < degree g(x), then f(x)/g(x) is called an improper rational function.

For example, rational function 1/(x²-4) can be rewritten as 1/4(x-2) -1/4(x+2) and rational function 3x/(x²+x-2) can be rewritten as 1/(x-1) + 2/(x+2). 

When to Use Partial Fractions

Partial fraction decomposition is applicable when integrating a rational function P(x)/Q(x)​, where:

• Degree of P(x) is less than the degree of Q(x): If not, perform polynomial long division first.
• Q(x) can be factored into linear and/or irreducible quadratic factors over the real numbers.

Integration by Partial Fraction Method

To evaluate the integral ∫[p(x)/q(x)] dx where p(x)/q(x) is in a proper rational fraction, we can factorize the denominator i.e., q(x) then using the following rational fraction cases we can write the integrand in a form of the sum of simpler rational functions including constant A, B, C, etc. Then values of  A, B, C, etc. can be obtained using various methods of algebra.

How to Integrate using Partial Fractions?

To integrate any rational function using Partial Fractions, we need to follow the following steps:

• Step 1: Factor the denominator given rational function into linear and quadratic factors.
• Step 2: Use the Partial Fraction formula to write the rational function as a sum of simpler fractions.
• Step 3: Determine the constants A, B, and C.
• Step 4: Integrate each partial fraction separately with appropriate methods to get the final integral.

Example: Integrate the following function using partial fractions: f(x) = (3x² + 2x + 1)/(x³ + x²)

Solution:

Step 1: Factor the denominator into linear and quadratic factors: x³ + x² = x²(x + 1)

Step 2: Write the rational function as a sum of simpler fractions.

f(x) = (3x² + 2x + 1)/[x²(x + 1)] = A/x + B/(x²) + C/(x+1)

Step 3: Determine the constants A, B, and C.

Multiplying both sides by the common denominator (x²(x + 1)), we get:

3x² + 2x + 1 = Ax(x+1) + B(x+1) + C(x²)

Substituting x = 0, x = -1, and x = infinity into the above equation, we get:

When x = 0, B = 1
When x = -1,  C = 2
When x = 1, A = 1

Solving the above equations simultaneously, we get:
A = 1, B = 1, C = 2

Step 4: Integrate each partial fraction using substitution.

Integrating A/x = 1/x, we get ln|x|
Integrating B/(x²) = 1/x², we get: -1/x
Integrating C/(x+1) = 2/(x+1), we get: @ ln|x+1|

Therefore, the final answer is:
∫f(x)dx = ln|x| - 1/x + 2 ln|x+1| + C, where C is the constant of integration.

Integration by Partial Fractions Examples

Example 1: Evaluate ∫(x - 1)/(x + 1)(x - 2) dx?

Solution:

Let (x - 1)/(x + 1)(x - 2) = A/(x + 1) + B/(x - 2)

Then, (x - 1) = A(x - 2) + B(x + 1) . . .(i)

Putting x = -1 in (i), we get A = 2/3

Putting x = 2 in (i), we get B = 1/3

Therefore,  

(x - 1)/(x + 1)(x - 2) = 2/3(x + 1) + 1/3(x - 2)
⇒ I = ∫(x - 1)/(x + 1)(x - 2) = 2/3∫dx/(x + 1) + 1/3∫dx/(x - 2)
⇒ I =  2/3log | x + 1 | + 1/3 log | x - 2 | + C

Example 2: Evaluate ∫dx/(x³ + x² + x + 1)?

Solution:

Let I = ∫dx/(x³ + x² + x + 1)

Now, 1/(x³ + x² + x + 1) = 1/[x²(x + 1) + (x + 1)] = 1/(x + 1)(x² + 1)

Let 1/(x + 1)(x² + 1) = A/(x + 1) + Bx + C/(x² + 1)  . . . (i)
⇒ 1 = A(x² + 1) + (Bx + C) (x + 1)

Putting x = -1 on both sides of (i), we get A = 1/2.

Comparing coefficients of x² on the both sides of (i), we get

A + B = 0 ⇒ B = -A = -1/2

Comparing coefficients of x on the both sides of (i), we get

B + C = 0 ⇒ C = -B = 1/2

Therefore, 1/(x + 1) (x² + 1) = 1/2(x + 1) + (-1/2x + 1/2)/(x² + 1)

Therefore, I = ∫dx/(x + 1) (x² + 1)
⇒ I = 1/2∫dx/(x + 1) - 1/2∫x/(x² + 1)dx + 1/2∫dx/(x² + 1)
⇒ I = 1/2∫dx/(x + 1) - 1/4∫2x/(x² + 1)dx + 1/2∫dx/(x² + 1)
⇒ I  = 1/2 log | x + 1 | - 1/4 log | x² + 1 | + 1/2 tan^-1x + C

Example 3: Evaluate ∫dx/x{6(log x)² + 7log x + 2}?

Solution:  

Putting log x = t and 1/x dx = dt, we get

I = ∫dx/x{6(log x)² + 7log x + 2} = ∫dt/(6t² + 7t + 2) = ∫dt/(2t + 1)(3t + 2)

Let 1/(2t + 1)(3t + 2) = A/(2t + 1) + B/(3t + 2)

Then, 1 ≡ A(3t + 2) + B(2t + 1)  . . . (i)

Putting t = -1/2 in (i), we get A = 2

Putting t = -2/3 in (i), we get B = -3

Therefore, 1/(2t + 1)(3t + 2) = 2/(2t + 1) + (-3)/(3t + 2)
⇒ I = ∫dt/(2t + 1)(3t + 2)
⇒ I = ∫2dt/(2t + 1) - ∫3dt/(3t - 2)
⇒ I = log | 2t + 1 | - log | 3t + 2 |
⇒ I = log | 2t + 1 |/log | 3t + 2 | + C
⇒ I = log | 2 log x + 1 | / log | 3 log x + 2 | + C

Example 4: Evaluate ∫dx/x(x⁴ + 1).

Solution:

We have

I = ∫dx/x(x⁴ + 1) = ∫x³/x⁴ (x⁴ + 1) dx [multiplying numerator and denominator by x³].

Putting x⁴ = t and 4x³dx = dt, we get
⇒ I  = 1/4∫dt/t(t + 1)
⇒ I = 1/4∫{1/t - 1/(t + 1)}dt   [by partial fraction]
⇒ I = 1/4∫1/t dt - 1/4∫1/(t + 1)dt
⇒ I = 1/4 log | t | - 1/4 log | t + 1 | + C
⇒ I = 1/4 log | x⁴ | - 1/4 log | x⁴ + 1 | + C
⇒ I =  (1/4 * 4) log | x | - 1/4 log | x⁴ + 1 | + C
⇒ I =  log | x | - 1/4 log | x⁴ + 1 | + C

Example 5: Evaluate ∫x²/(x² + 2)(x² + 3)dx?

Solution: 

Let x²/(x² + 2) (x² + 3) = y/(y + 2)(y + 3) where x² = y.

Let y/(y + 2) (y + 3) = A/(y + 2) + B/(y + 3)

⇒ y ≡ A(y + 3) + B/(y + 2)  . . . (i)

Putting y = -2 on the both sides of (i), we get A = -2.

Putting y = -3 on the both sides of (i), we get B = 3.

Therefore, y/(y + 2) (y + 3) = -2/(y + 2) + 3/(y + 3)
⇒  x²/(x² + 2) (x² + 3) = -2/(x² + 2) + 3/(x² + 3)
⇒  ∫x²/(x² + 2) (x² + 3) = -2∫dx/(x² + 2) + 3∫dx/(x² + 3)
⇒  ∫x²/(x² + 2) (x² + 3) = -2/√2tan^-1(x/√2) + 3/√3tan^-1(x/√3) + C
⇒  ∫x²/(x² + 2) (x² + 3) = -√2tan^-1(x/√2) + √3tan^-1(x/√3) + C

Integration by Partial Fractions Practice Problems

Solve these Integrals using integration by Partial Fractions:

1.

2.

3.

4.

5.

Related Articles:

• Integrals
• Indefinite Integrals
• Integration Formulas
• Integration by Parts