Integration by Partial Fractions is one of the methods of integration, which is used to find the integral of the 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 polynomial functions such functions. that g(x) ≠ 0 then f(x)/g(x) is called Rational Functions. 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]. In this article, we will learn about the methods of Integration by Partial Fractions with detailed solved examples.

- What is Integration by Partial Fractions?
- Integration by Partial Fraction Method
- How to Integrate using Partial Fractions?
- Solved Examples of Integration by Partial Fractions
- FAQs on Integration by Partial Fractions
- Q1: What is Integration by Partial Fractions?
- Q2: When is Integration by Partial Fractions Used?
- Q3: What is a Rational Function?
- Q4: What is a Partial Fraction Decomposition?
- Q5: What is a Linear Factor?
- Q6: What is a Quadratic Factor?
- Q7: Can every Rational Function be decomposed into Partial Fractions?

## What is Integration by Partial Fractions?

Integration by Partial Fractions is a method for decomposing any improper or complex proper rational function into a sum of at least two proper rational functions. In other words, in this method, we decompose complex hard to integrate rational functions into simple easy to integrate rational functions.

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).

## 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) = (3×2 + 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) = (3×2 + 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:

3×2 + 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 = 1Solving the above equations simultaneously, we get:

A = 1, B = 1, C = 2Step 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.

Also, Read

## Solved Examples of Integration by Partial Fractions

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/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 3: 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 4: 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

Example 5: 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

## FAQs on Integration by Partial Fractions

### Q1: What is Integration by Partial Fractions?

Answer:

Integration by Partial Fractions is a method of integration used to integrate the rational function with a complex denominator and numerator.

### Q2: When is Integration by Partial Fractions Used?

Answer:

A partial Fraction is used when a rational function seems complicated at glance to integrate, so we use a partial fraction to covert the complex rational function into a sum of simple rational functions where mostly numerator is a real number.

### Q3: What is a Rational Function?

Answer:

For two polynomial functions f(x) and g(x) rational function is defined as f(x)/g(x) where g(x) can’t be 0.

### Q4: What is a Partial Fraction Decomposition?

Answer:

Partial Fraction decomposition is the method of simplifying complex rational functions into simple rational functions.

### Q5: What is a Linear Factor?

Answer:

A factor of form ax+b is called linear factor where a and b are real numbers.

### Q6: What is a Quadratic Factor?

Answer:

A factor of form ax2+bx+c is called quadratic factor where a, b, and c are real numbers.

### Q7: Can every Rational Function be decomposed into Partial Fractions?

Answer:

Yes, every rational function can be decomposed into partial fractions if there exist only linear and quadratic factors of the denominator of the rational function.