HOW TO INTEGRATE A RATIONAL FUNCTION LINEAR IN THE NUMERATOR

To know the formulas used in integration, please visit the page "Integration Formulas for Class 12".

Question 1 :

Evaluate the following with respect to "x".

(3x + 1) / (2x² - 2x + 3)

Solution :

 ∫ (3x + 1) / (2x² - 2x + 3)dx

(3x + 1)   =   A(d/dx) (2x² - 2x + 3) + B

3x + 1  =  A (4x - 2) + B  ----(1)

Equating the coefficients of x.

3  =  4A

A  =  3/4

Equating constant terms

1  =  -2A + B

1  =  -2(3/4) + B

1  =  -3/2 + B

B  =  1 + (3/2)  ===>  B  =  5/2

Applying the values of A and B in (1)

3x + 1  =  (3/4) (4x - 2) + (5/2) 

By dividing each term by (2x² - 2x + 3), we get

∫ (3x + 1) / (2x² - 2x + 3)dx 

  = (3/4)∫(4x-2)/(2x²-2x+3) dx+(5/2)∫1/(2x²-2x+3)dx

  =  (3/4)log (2x²-2x+3)+(5/2)∫1/(2x²-2x+3)dx

2x²-2x+3 = 2[x² - x + (3/2)]

=  2[x² - 2x(1/2) + (1/2)² - (1/2)² + (3/2)]

=  2[(x-(1/2))²-(1/4) + (3/2)]

=  2 [(x-(1/2))² + (√(5/2))²]

=  (3/4)log (2x²-2x+3)+(5/2)∫1/2 [((2x-1)/2)²+√(5/2)²]dx

=  (3/4)log (2x²-2x+3)+(√5/2)tan^-1[(2x-1)/√5]

Question 2 :

Evaluate the following with respect to "x".

(2x + 1) / √(9 + 4x - x²)

Solution :

 ∫(2x + 1) / √(9 + 4x - x²) dx

(2x + 1)   =   A(d/dx) (9 + 4x - x²) + B

2x + 1  =  A (-2x + 4) + B  ----(1)

Equating the coefficients of x.

2  =  -2A

A  =  -1

Equating constant terms

1  =  4A + B

1  =  4(-1) + B

1  =  -4 + B

B  =  1 + 4 ===>  B  =  5

Applying the values of A and B in (1)

2x + 1  =  -1 (-2x + 4) + 5 

By dividing each term by (2x² - 2x + 3), we get

∫(2x + 1) / √(9 + 4x - x²) dx

  = -1∫(-2x+4)/√(9 + 4x - x²) dx + 5 ∫1/√(9 + 4x - x²)dx

  =  -2 log √(9 + 4x - x²) + 5 ∫1/√(9 + 4x - x²)dx

√(9 + 4x - x²) = -[x² - 4x - 9]

  =  [x² - 2x(2) + 2² - 2² - 9]

  =  [(x - 2)² - 13]

  =  [(x - 2)² - (√13)²]

  =  -2 log √(9 + 4x - x²) + 5 ∫1/√ [(x - 2)² - (√13)²]dx

  =  -2 log √(9 + 4x - x²) + 5 sin^-1 [(x - 2)/√13]

  =  + 5 sin^-1 [(x - 2)/√13] - 2 log √(9 + 4x - x²) + c

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