HOW TO INTEGRATE QUADRATIC EQUATION IN THE SQUARE ROOT

To know the formulas used in integration, please visit the page "Integration Formulas for Class 12".

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

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

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

Question 1 :

Evaluate the following with respect to "x".

 √(x2 + 2x + 10)

Solution :

 ∫ √(x2 + 2x + 10) dx 

x2 + 2x + 10  =  x2 + 2x(1) + 12 - 12 + 10 

=  (x + 1)2 - 1 + 10 

=  (x + 1)2 + 9 

=  (x + 1)2 + 32

 ∫ √(x2 + 2x + 10) dx  =   ∫√[(x + 1)2 + 32] dx 

=  ((x+1)/2)√(x2+2x+10) + (9/2)log (x+√(x2+2x+10)) + c

Question 2 :

Evaluate the following with respect to "x".

 √(x2 - 2x - 3)

Solution :

 √(x2 - 2x - 3) dx 

x2 - 2x - 3  =  x2 - 2x(1) + 12 - 12 -3

=  (x - 1)2 - 1 - 3 

=  (x - 1)2 - 4 

=  (x - 1)2 - 22

 ∫  √(x2 - 2x - 3) dx  =   ∫√[(x - 1)2 - 22] dx 

=  ((x-1)/2)√(x2 - 2x - 3) - (4/2)log (x-1+√(x2-2x-3)) + c

=  ((x-1)/2)√(x2 - 2x - 3) - 2 log (x-1 + √(x2 - 2x - 3)) + c

Question 3 :

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

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