INDEFINITE INTEGRALS WITH SQUARE ROOTS

Question 1 :

Evaluate the following with respect to "x".

 √(6 - x)(x - 4)

Solution :

 √(6 - x)(x - 4) dx 

(6 - x)(x - 4)  =  6x - 24 - x² + 4x

  =  -x² + 10x - 24

  =  -[x² - 10x + 24]

=  -[x² - 2x(5) + 5² - 5² + 24]

=  -[(x-5)² - 25 + 24]

=  -[(x-5)² - 1]

=  1² - (x-5)²

∫(√a²-x²) dx = (x/2)(√a²-x²)+(a²/2) sin^-1(x/a) + c

 ∫√(-x² + 10x - 24) dx  =   ∫√[1² + (x-5)²] dx 

=  ((x-5)/2)√(-x² + 10x - 24)+(1/2)sin^-1((x-5)/1) + c

Question 2 :

Evaluate the following with respect to "x".

 √[9 - (2x + 5)²]

Solution :

 ∫√[9 - (2x + 5)²] dx 

√[9 - (2x + 5)²] =  √[3² - (2x + 5)²] dx

∫(√a²-x²)dx=(x/2)(√a²-x²)+(a²/2) sin^-1(x/a)+c

=  (2x + 5)/2√[9 - (2x + 5)²] + (9/2) sin^-1[(2x + 5)/3] + c

Question 3 :

Evaluate the following with respect to "x".

 √[81 + (2x + 1)²]

Solution :

 ∫√[81 + (2x + 1)²] dx 

√[81 + (2x + 1)²]  =  √[9² + (2x + 1)²] dx

∫(√x²+a²) dx = (x/2)(√x²+a²)+(a²/2) log (x+√(x²+a²)+c

=  ((2x+1)/3)√[9²+(2x+1)²]+(9/2) log (2x+1)+√[9²+ (2x+1)²] 

=  ((2x+1)/3)√[81 + (2x + 1)²]+(9/2) log (2x+1)+√[81 + (2x + 1)²] + c

Question 4 :

Evaluate the following with respect to "x".

 √[(x + 1)² - 4]

Solution :

 ∫ √[(x + 1)² - 4] dx

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

∫(√x²-a²)dx=(x/2)(√a²-x²)-(a²/2)log(x+ √a²-x²) + c

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