Question

Integrate the following with respect to x:

`(3x + 1)/(2x^2 - 2x + 3)`

Answer 1

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

3x + 1 = A(4x – 2) + B

3x + 1 = 4Ax – 2A + B

4A = 3

⇒ A = `3/4`

– 2A + B = 1

`- 2 xx 3/4 + "B"` = 1

`3/2 + "B"` = 1

B = `1 + 3/2 = 5/2`

B = `5/2`

3x + 1 = `3/4 (4x - 2) + 5/2`

`int (3x + 1)/(2x^2 - 2x + 3)  "d"x = int (3/4 (4x - 2) + 5/2)/(2x^2 - 2x + 3)  "d"x`

= `3/4 int (4x - 2)/(2x^2 - 2x + 3) + 5/2 int ("d"x)/(2x^2 - 2x + 3)`

Put `2x^2 - 2x + 3` = t

`(4x - 2) "d"x` = dt

= `3/4 int "dt"/"t" + 5/2 int ("d"x)/(2(x^2 - x + 3/2))`

= `3/4 log |"t"| + 5/4 int ("d"x)/((x - 1/2)^2 - (1/2)^2 + 3/2)`

= `3/4 log |2x^2 - 2x + 3| + 5/4 int ("d"x)/((x - 1/2)^2 + - 1/4 + 3/2)`

= `3/4 log |2x^2 - 2x + 3| + 5/4 int ("d"x)/((x - 1/2)^2 + (6 - 1)/4`

= `3/4 log |2x^2 - 2x + 3| + 5/4 int ("d"x)/((x - 1/2)^2 + 5/4)`

= `3/4 log |2x^2 - 2x + 3| + 5/4 int ("d"x)/((x - 1/2)^2 + (sqrt(5)/2)^2`

`int ("d"x)/(x^2 + "a"^2) = 1/"a" tan^-1 (x/"a") + "c"`

= `3/4 log |2x^2 - 2x + 3| + 5/4 xx tan^-1 ((x - 1/2)/(sqrt(5)/2)) + "c"`

= `3/4 log |2x^2 - 2x + 3| + 5/4 xx 1/(sqrt(5)/2) tan^-1 ((2x - 1)/sqrt(5)) + "c"`

= `3/4 log |2x^2 - 2x + 3| + 5/4 xx 2/sqrt(5) tan^-1 ((2x - 1)/sqrt(5)) + "c"`

= `3/4 log |2x^2 - 2x + 3| + sqrt(5)/2 tan^-1 ((2x - 1)/sqrt(5)) + "c"`