Question

Integrate the following with respect to x:

`(2x + 3)/sqrt(x^2 + 4x + 1)`

Answer 1

Let 2x + 3 = `"A" "d"/("d"x) (x^2 + 4x + 1) + "B"`

2x + 3 = A(2x + 3) + B

2A = 2

⇒ A = 1

4A + B = 3

4 × 1 + B = 3

⇒ B = 3 – 4

B =  – 1

(2x + 3) = (2x + 4) – 1

`int (2x + 3)/sqrt(x^2 + 4x + 1)  "d"x = int ((2x + 4) - 1)/sqrt(x^2 + 4x + 1)  "d"x`

= `int ((2x + 4))/sqrt(x^2 + 4x + 1)  "d"x - int ("d"x)/sqrt(x^2 + 4x + 1)`

Put x² + 4x + 1 = t

(2x + 4) dx = dt

= `intt "dt"/sqrt("t") - int ("d"x)/sqrt((x + 2)^2 - 2^2 + 1)`

= `int "t"^(- 1/2)  "dt" - int ("d"x)/sqrt((x + 2)^2 - 3)`

= `("t"^(- 1/2 + 1))/(- 1/2 + 1) - int ("d"x)/sqrt((x + 2)^2 - (sqrt(3))^2`

Put x + 2 = u

dx = dt

= `"t"^(1/2)/(1/2) - int "du"/sqrt("u"^2 - (sqrt(3))^2`

= `2sqrt("t") - log |"u" + sqrt("u"^2 - (sqrt(3))^2)| + "c"`

= `2 sqrt(x^2 + 4x + 1) - log |(x + 2) + sqrt((x + 2)^2 - (sqrt(3))^2)| + "c"`

= `2 sqrt(x^2 + 4x + 1) - log|(x + 2) + sqrt(x^2 + 4x + 4 - 3)| + "c"`

`int (2x + 3)/sqrt(x^2 + 4x + 1)  "d"x = 2 sqrt(x^2 + 4x + 1) - log|(x + 2) + sqrt(x^2 + 4x + 1)| + "c"`