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) - 71/(x2+4x-12) dx

x2+4x-12  =  x+ 2x(2) + 22 - 22 -12

 =  (x + 2)2 + 4 - 12

 =  (x + 2)2 - 8

 =  (x + 2)2 - 42

  =  log (x2+4x-12) - 71/[(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 + x2dx 

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

  =  log (2 + 2x + x2) - 71/(2 + 2x + x2) dx

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

 =  (x + 1)2 - 1 + 2

 =  (x + 1)2 + 1

  =  (5/2)log (2 + 2x + x2) - 71/[(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

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