INTEGRATION OF LINEAR IN NUMERATOR AND QUADRATIC IN THE DENOMINATOR

Question 1 :

Evaluate the following with respect to "x".

(2x - 3) / (x² + 4x - 12)

Solution :

 ∫(2x - 3) / (x² + 4x - 12) dx

(2x - 3)   =   A(d/dx) (x² + 4x - 12) + B

2x - 3  =  A (2x + 4) + B  ----(1)

Equating the coefficients of x.

2  =  2A

A  =  1

Equating constant terms

-3  =  4A + B

-3  =  4(1) + B

-3  =  4 + B

B  =  -3 - 4  ===>  B  =  -7

Applying the values of A and B in (1)

2x - 3  =  1 (2x + 4) - 7

By dividing each term by , we get

∫(2x-3)/(x²+4x-12) dx 

  = ∫(2x-4)/(x²+4x-12) dx - 7 ∫1/(x²+4x-12) dx

  =  log (x²+4x-12) - 7∫1/(x²+4x-12) dx

x²+4x-12  =  x² + 2x(2) + 2² - 2² -12

 =  (x + 2)² + 4 - 12

 =  (x + 2)² - 8

 =  (x + 2)² - 4²

  =  log (x²+4x-12) - 7∫1/[(x + 2)² - 4²] dx

  =  log (x²+4x-12) - 7 [1/2(4) log(x + 2 - 4) / (x + 2 + 4)]

  =  log (x²+4x-12) - (7/8) [log(x-2) / (x + 6)] + c

Question 2 :

Evaluate the following with respect to "x".

(5x - 2) / (2 + 2x + x²)

Solution :

 ∫(5x - 2) / (2 + 2x + x²) dx

(5x - 2)   =   A(d/dx) (2 + 2x + x²) + B

5x - 2  =  A (2 + 2x) + B  ----(1)

Equating the coefficients of x.

5  =  2A

A  =  5/2

Equating constant terms

-2  =  2A + B

-2  =  2(5/2) + B

-2  =  5 + B

B  =  -2 - 5  ===>  B  =  -7

Applying the values of A and B in (1)

5x - 2  =  (5/2) (2 + 2x) - 7

By dividing each term by , we get

∫(5x - 2) / (2 + 2x + x²) dx 

  =  (5/2) ∫(2x + 2) / (2 + 2x + x²) - 7 ∫1/(2 + 2x + x²) dx

  =  log (2 + 2x + x²) - 7∫1/(2 + 2x + x²) dx

2 + 2x + x²  =  x² + 2x(1) + 1² - 1² + 2

 =  (x + 1)² - 1 + 2

 =  (x + 1)² + 1

  =  (5/2)log (2 + 2x + x²) - 7∫1/[(x + 1)² + 1] dx

  =  (5/2) log (2 + 2x + x²) - 7 tan^-1 (x + 1) + c

  =  (5/2) log (2 + 2x + x²)  - 7 tan^-1 (x + 1) + c

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.