Mathematics Integration by Partial Fractions and Integration by Parts FOR CBSE-NCERT
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### Topic covered

color{red} {star} Integration by Partial Fractions
color{red} {star} Integration by Parts

### Integration by Partial Fractions

color{green}"Steps 1 :" Factor the denominator.

color{green}"Steps 2 :" Break up the fraction on the right into a sum of fractions, where each factor of the denominator in Step 1 becomes the denominator of a separate fraction.

color{green}"Steps 3 :" Multiply both sides of this equation by the denominator of the left side.

color{green}"Steps 4 :" Take the roots of the linear factors and plug them one at a time into x in the equation from Step 3, and solve for the unknowns.

color{green}"Steps 5 :" Plug these results into the A and B in the equation from Step 2.

color{green}"Steps 6 :" Split up the original integral into the partial fractions from Step 5 and integrate.

The following Table indicates the types of simpler partial fractions that are to be associated with various kind of rational functions.

Q 3125678561

Find  ∫ ( dx)/( (x+1) (x+2) )

Class 12 Chapter 7 Example 11
Solution:

The integrand is a proper rational function. Therefore, by using the form of
partial fraction [Table 7.2 (i)], we write

1/( ( x+1) (x+2) ) = A/(x+1) +B/(x+2) .......... (1)

where, real numbers A and B are to be determined suitably. This gives

1 = A (x + 2) + B (x + 1).

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

A + B = 0

and 2A + B = 1

Solving these equations, we get A =1 and B = – 1.

Thus, the integrand is given by

1/((x+1)(x+2)) = 1/(x+1) + (-1)/(x+2)

Therefore,  ∫ (dx)/( (x+1)(x+2) ) = ∫ (dx)/(x+1) - ∫ (dx)/(x+2)

= log | x+1 | - log | x+2 | +C

= log | (x+1)/(x+2) | +C
Q 3155678564

Find ∫ (x^2 +1)/(x^2 - 5x + 6 ) dx
Class 12 Chapter 7 Example 12
Solution:

Here the integrand  (x^2 +1 )/(x^2 -5x + 6 ) is not proper rational function, so we divide
x^2 + 1 by x^2 – 5x + 6 and find that

 (x^2 +1)/(x^2 - 5x + 6 ) =1+ (5x-5)/(x^2 - 5x + 6) =1 + (5x -5)/( (x-2) (x-3) )

Let  (5x-5)/( (x-2)(x-3) ) = A/(x-2) +B/(x-3)

So that 5x – 5 = A (x – 3) + B (x – 2)
Equating the coefficients of x and constant terms on both sides, we get A + B = 5
and 3A + 2B = 5. Solving these equations, we get A = – 5 and B = 10

Thus  (x^2 +1)/(x^2 - 5x + 6 ) = 1- 5/(x-2) +10/(x-3)

Therefore,  ∫ (x^2+1)/(x^2 - 5x + 6 ) dx = ∫ dx -5 ∫ 1/(x-2) dx +10 ∫ (dx)/(x-3)

 = x-5 log |x-2 | + 10 log | x-3 | +C
Q 3175678566

Find  ∫ (3x-2)/( (x+1)^2 (x+3) ) dx

Class 12 Chapter 7 Example 13
Solution:

The integrand is of the type as given in Table 7.2 (4). We write

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

So that 3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)^2

= A (x^2 + 4x + 3) + B (x + 3) + C (x^2 + 2x + 1 )

Comparing coefficient of x2, x and constant term on both sides, we get

A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2. Solving these equations, we get

A = 11/4 , B = (-5)/2 and C = (-11)/4 . Thus the integrand is given by

 (3x-2)/( (x+1)^2 (x+3) ) = 11/(4 (x+1) ) - 5/(2 (x+1)^2 ) - 11/(4 (x+3) )

Therefore,  ∫ (3x -2)/( (x+1)^2 (x+3) ) = 11/4 ∫ (dx)/(x+1) - 5/2 ∫ (dx)/((x+1)^2 ) = 11/4 ∫ (dx)/(x+3)

= 11/4 log | x+1 | + 5/(2 (x+1) ) -11/4 log | x+3 | +C

= 11/4 log | (x+1)/(x+3) | + 5/(2 (x+1) ) +C
Q 3115678569

Find  ∫ (x^2 )/( (x^2 +1 ) (x^2 +4) ) dx

Class 12 Chapter 7 Example 14
Solution:

Consider  (x^2 )/( (x^2+1) ( x^2 +4) ) and put x^2 = y .

Then  x^2/( (x^2+1) (x^2+4) ) = y/( (y+1)( y+4) )

Write  y/( (y+1)(y+4) ) = A/(y+1) +B/(y+4)

So that y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides, we get A + B = 1
and 4A + B = 0, which give

A = -1/3 and B = 4/3

Thus,  (x^2)/( (x^2+1) ( x^2+4) ) = -1/(3 (x^2+1) ) + 4/(3 (x^2 +4) )

Therefore,  ∫ (x^2 dx)/( (x^2 +1) (x^2 +4) ) = -1/3 ∫ (dx)/(x^2+1) + 4/3 ∫ (dx)/(x^2 +4)

 = -1/3 tan^(-1) x +4/3 xx 1/2 tan^(-1) x/2 +C

 = -1/3 tan^(-1) x + 2/3tan^(-1) x/2 +C

In the above example, the substitution was made only for the partial fraction part
and not for the integration part. Now, we consider an example, where the integration
involves a combination of the substitution method and the partial fraction method.
Q 3155778664

Find  ∫ ( ( 3 sin φ -2 ) cos φ )/( 5- cos^2 φ - 4 sin φ) d φ
Class 12 Chapter 7 Example 15
Solution:

Let y = sinφ
Then dy = cosφ dφ

Therefore,  ∫ ( ( 3 sin φ -2 ) cos φ )/(5 - cos^2 φ -4 sin φ) d φ = int ( ( 3y -2) dy)/( 5- (1-y^2 ) -4y )

= ∫ (3y -2)/(y^2 -4y +4) dy

 = ∫ (3y -2)/( (y-2)^2 ) =I (say )

Now, we write  (3y -2)/( (y-2)^2 ) = A/(y-2) +B/( (y-2)^2 ) [by Table 7.2 (2)]

Therefore, 3y – 2 = A (y – 2) + B
Comparing the coefficients of y and constant term, we get A = 3 and B – 2A = – 2,
which gives A = 3 and B = 4.
Therefore, the required integral is given by

I = ∫ [ 3/(y-2) + 4/( (y-2)^2) ] dy = 3 ∫ (dy)/( y-2) +4 ∫ (dy)/( (y-2)^2 )

= 3 log | y-2 | +4 ( -1/(y-2) ) +C

= 3 log | sin φ -2 | + 4/(2- sin φ) +C

 =3 log (2- sin φ ) + 4/(2- sin φ) +C (since, 2 – sinφ is always positive)
Q 3185778667

Find  ∫ (x^2 +x + 1 dx)/( (x+2) (x^2 +1 ) )
Class 12 Chapter 7 Example 16
Solution:

The integrand is a proper rational function. Decompose the rational function
into partial fraction [Table 2.2(5)]. Write

 (x^2 + x+1 )/( (x^2 +1) (x+2) ) =A/(x+2) + (Bx+C)/( (x^2 +1 ) )

Therefore, x^2 + x + 1 = A (x^2 + 1) + (Bx + C) (x + 2)

Equating the coefficients of x^2, x and of constant term of both sides, we get
A + B =1, 2B + C = 1 and A + 2C = 1. Solving these equations, we get

A = 3/5 , B =2/5 and C = 1/5

Thus, the integrand is given by

 (x^2+ x +1)/( (x^2 +1) (x+2) ) = 3/(5 (x+2) ) + (2/5 x + 1/5 )/(x^2 +1) = 3/(5 (x+2) ) +1/5 ( (2x+1)/(x^2 +1) )

Therefore,  ∫ ( x^2 + x +1 )/( (x^2 +1 ) (x+2) ) dx =3/5 ∫ (dx)/(x+2) +1/5 ∫ (2x)/(x^2 +1) dx +1/5 ∫ 1/(x^2 +1) dx

 =3/5 log | x+2| + 1/5 log | x^2 +1 | + 1/5 tan^(-1) x +C