Question

Evaluate : ∫(x - 3)√(x^2 + 3x - 18) dx

Answer 1

Step-by-step explanation:

Given Evaluate : ∫(x - 3)√(x^2 + 3x - 18) dx

• Now we can write x – 3 as P + Q d/dx (x^2 + 3x – 18)

•    So x – 3 = P + Q (2x + 3)
• Now comparing the coefficient of x we get  
•                                                              2 Q = 1
•                                                              Or Q = ½  
• Also constant term will be
•                                             So – 3 = P + 3 Q
•                                            Or – 3 = P + 3(1/2)  
•                                               So P = - 9/2
• Now substituting the values of P and Q we have
•                 So ∫(- 9/2 + ½ (2x + 3) ) √x^2 + 3x – 18 dx
•           So – 9/2 ∫√x^2 + 3x – 18 + 1/2 ∫(2x + 3) √x^2 + 3x – 18 dx
• [Now it is a quadratic equation and we can write this as
•           Now coefficient of x is 3, multiply 3 by ½ we get 3/2, so (3/2)^2 = 9/4
•        So (x^2 + 3x + 9/4) – 18 – 9/4
•          So (x + 3/2)^2 – (9/2)^2]
• So – 9/2 ∫√ (x + 3/2)^2 – (9/2)^2 dx + ½ ∫ √t dt + ½ [ t^3/2 / 3/2] + C
• Now we have ∫ √x^2 – a^2 = x/2 √x^2 – a^2 – a^2/2 log (x + √x^2 - a^2)
•        So – 9/2 [ ½ (x + 3/2) √x^2 + 3x – 18 – 81 / 4 log [ (x + 3/2) + √x^2 + 3x – 18] + 1/3 (x^2 + 3x – 18)^3/2
•  So – 9/2 (2x + 3 / 4 √x^2 + 3x – 18 – 81 / 8 log (2x + 3 / 2 + √x^2 + 3x – 18) + 1/3 (x^2 + 3x – 18)^3/2 + C

Reference link will be

https://brainly.in/question/14791355

Answer 2

Answer:

Step-by-step explanation:

Given Evaluate : ∫(x - 3)√(x^2 + 3x - 18) dx

Now we can write x – 3 as P + Q d/dx (x^2 + 3x – 18)

   So x – 3 = P + Q (2x + 3)

Now comparing the coefficient of x we get  

                                                             2 Q = 1

                                                             Or Q = ½  

Also constant term will be

                                            So – 3 = P + 3 Q

                                           Or – 3 = P + 3(1/2)  

                                              So P = - 9/2

Now substituting the values of P and Q we have

                So ∫(- 9/2 + ½ (2x + 3) ) √x^2 + 3x – 18 dx

          So – 9/2 ∫√x^2 + 3x – 18 + 1/2 ∫(2x + 3) √x^2 + 3x – 18 dx

[Now it is a quadratic equation and we can write this as

          Now coefficient of x is 3, multiply 3 by ½ we get 3/2, so (3/2)^2 = 9/4

       So (x^2 + 3x + 9/4) – 18 – 9/4

         So (x + 3/2)^2 – (9/2)^2]

So – 9/2 ∫√ (x + 3/2)^2 – (9/2)^2 dx + ½ ∫ √t dt + ½ [ t^3/2 / 3/2] + C

Now we have ∫ √x^2 – a^2 = x/2 √x^2 – a^2 – a^2/2 log (x + √x^2 - a^2)

       So – 9/2 [ ½ (x + 3/2) √x^2 + 3x – 18 – 81 / 4 log [ (x + 3/2) + √x^2 + 3x – 18] + 1/3 (x^2 + 3x – 18)^3/2

 So – 9/2 (2x + 3 / 4 √x^2 + 3x – 18 – 81 / 8 log (2x + 3 / 2 + √x^2 + 3x – 18) + 1/3 (x^2 + 3x – 18)^3/2 + C