INTEGRATION OF  RATIONAL FUNCTION WITH SQUARE ROOT IN DENOMINATOR

Question 1 :

Evaluate the following with respect to "x".

(x + 2) / √(x2 - 1)

Solution :

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

(x/√(x2 - 1)) dx 

x2 - 1  =  t

2x dx  =  dt

x dx  =  dt/2

(x/√(x2 - 1)) dx  =  ∫(1/√t) dt

  =  t-1/2 (dt/2)

  =  (1/2) t1/2/(1/2)

  =  (x2 - 1)   ---(1)

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

(1) + (2)

  =  (x2 - 1) + 2 log(x + √(x2 - 1)) + c

Question 2 :

Evaluate the following with respect to "x".

(2x + 3) / √(x2 + 4x + 1)

Solution :

 ∫(2x + 3) / √(x+ 4x + 1) dx

(2x + 3)   =   A(d/dx) (x+ 4x + 1) + 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  =  -1

Applying the values of A and B in (1)

2x + 3  =  1 (2x + 4) - 1

By dividing each term by √(x+ 4x + 1), we get

(2x + 1) / √(x+ 4x + 1) dx

  =  (2x+4)/√(x+ 4x + 1) dx - 1 1/√(x+ 4x + 1)dx

  =  2 log (x+ 4x + 1) - 1 1/(x+ 4x + 1)dx

(x+ 4x + 1)  =   [x+ 2x(2) + 22 - 22 + 1]

  =  [(x + 2)- 3]

  =  [(x + 2)32]

  =  2 log (x+ 4x + 1) - 1 1/[(x + 2)32] dx

  =  2 log (x+ 4x + 1) - log (x + 2) + (x+ 4x + 1) + c

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