Question

Solve the following : `int_1^2 (5x^2)/(x^2 + 4x + 3)*dx`

Answer 1

Let I = `int_1^2 (5x^2)/(x^2 + 4x + 3)*dx`

= `5 int_1^2 x^2/(x^2 + 4x + 3)*dx`
Dividing numerator by denominator, we get

                        1               
`x^2 + 4x + 3)x^2`
                      x² + 4x + 3
                      –      –      –   
                             – 4x – 3

∴ I = `5 int_1^2(1 - (4x + 3)/(x^2 + 4x + 3))*dx`

= `5 int_1^2 1*dx - 5 int_1^2 (4x + 3)/(x^2 + 4x + 3)*dx`

= `5 int_1^2 1*dx - 5 int_1^2 (4x + 3)/((x + 3)(x + 1))*dx`

Let `(4x + 3)/((x + 3)(x + 1)) = "A"/(x + 3) + "B"/(x + 1)`     ...(i)

∴ 4x + 3 =A(x + 1) + B(x + 3)    ...(ii)
Putting x = – 1 in (ii), we get
– 4 + 3 = A (– 1 + 1) + B(– 1 + 3)
∴ – 1 = 2B
∴ B = `-(1)/(2)`
Putting x = –3 in (ii), we get
– 12 + 3 = A(– 3 + 1) + B(– 3 + 3)
∴ – 9 = – 2A
∴ A = `(9)/(2)`
From (i), we get

`(4x + 3)/((x + 3(x + 1))`

= `(9/2)/(x + 3) + ((-1/2))/(x + 1)`

∴ I = `5 int_1^2 1*dx - 5 int_1^2[(9/2)/(x + 3) + ((-1/2))/(x + 1)]*dx`

= `5[x]_1^2 - 5[9/2 int_1^2 (1)/(x + 3)*dx - (1)/(2) int_1^2 (1)/(x + 1)*dx]`

= `5 (2 - 1) - 5{9/2 [log |x + 3|]_1^2  - (1)/(2)[log|x + 1|]_1^2}`

= `5 - 5[9/2(log 5 - log4)  - (1)/(2)(log 3 - log 2)]`

= `5 - (5)/(2)[9 (log 5 - log2^2) - (log 3 - log 2)]`

= `5 - (5)/(2) [9 (log 5 - 2 log 2)  log 3 + log 2]`

= `5 - (5)/(2) (- log 3 - 17 log 2 + 9 log 5)`

∴ I  = `5 + (1)/(2) (5 log 3 + 85 log 2 - 45 log 5)`.