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प्रश्न
Integrate the following with respect to x:
`(3x + 1)/(2x^2 - 2x + 3)`
उत्तर
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"`
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