Question

Integrate the following with respect to x:

`(5x - 2)/(2 + 2x + x^2)`

Answer 1

Let 5x – 2 = `"A" "d"/("d"x) (x^2 + 2x + 2) + "B"`

5x – 2 = A(2x + 2) + B

5x – 2 = 2Ax + 2A + B

2A = 5

⇒ A = `5/2`

2A + B = – 2

`2 xx 5/2 + "B"` = – 2

⇒ B = – 2 – 5 = – 7

5x – 2 = `- 5/2 (2x + 2) - 7`

`int (5x - 2)/(x^2 + 2x + 2)  "d"x = int (5/2 (2x + 2) - 7)/(x^2 + 2x + 2)  "d"x`

= `5/2 int (2x + 2)/(x^2 + 2x + 2)  "d"x - 7 int ("d"x)/(x^2 + 2x + 2)`

Put x² + 2x + 12 = t

(2x + 2)dx = dt

= `5/2 int "dt"/"t" - 7 int ("d"x)/((x + 1)^2 - 1^2 + 2)`

= `5/2 log |"t"| - 7 int ("d"x)/((x + 1)^2 + 1)`

= `5/2 log |x^2 + 2x + 2| - 7 int ("d"x)/((x + 1)^2 + 1^2)`

Put x + 1 = u

dx = du

= `5/2 log |x^2 + 2x + 2| - 7 int "du"/("u"^2 + 1^2)`

= `5/2 log |x^2 + 2x + 2| - 7 xx 1/1 tan^-1 ("u"/1) + "c"`

= `5/2 log|x^2 + 2x + 2| - 7 tan^-1 (x + 1) + "c"`