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) t^1/2/(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) / √(x² + 4x + 1) dx

(2x + 3)   =   A(d/dx) (x² + 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 √(x² + 4x + 1), we get

∫(2x + 1) / √(x² + 4x + 1) dx

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

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

√(x² + 4x + 1)  =   [x² + 2x(2) + 2² - 2² + 1]

  =  [(x + 2)² - 3]

  =  [(x + 2)² - √3²]

  =  2 log √(x² + 4x + 1) - 1 ∫1/√[(x + 2)² - √3²] dx

  =  2 log √(x² + 4x + 1) - log (x + 2) + √(x² + 4x + 1) + c

Related topics

• Indefinite Integral Using Properties
• Integration of Rational Functions with Quadratic Denominator
• Integration of Rational Functions With Square Roots
• Integration of Linear in Numerator and Quadratic in the Denominator
• How to Integrate a Rational Function Linear in the Numerator
• Integration of Rational Function With Square Root in Denominator

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