ज्ञात करें `int (2x+1)/(2x^2 +4x -3) dx`

Options

A. `I =1/2 log| 2x^2 +4x-3| - (sqrt2) /(4)log|(x+1-sqrt(5/2))/(x+1+sqrt(5/2))|+c`
B. `I =1/2 log| 2x^2 +4x-3| - (sqrt2) /(sqrt5)log|(x+1-sqrt(5/2))/(x+1+sqrt(5/2))|+c`
C. `I =1/2 log| 2x^2 +4x-3| - (sqrt2) /(4sqrt5)log|(x+1-sqrt(5/2))/(x+1+sqrt(5/2))|+c`
D. `I =1/2 log| 2x^2 +4x-3| - (1) /(4sqrt5)log|(x+1-sqrt(5/2))/(x+1+sqrt(5/2))|+c`

Answer

C

Explanation

माना कि `int (2x+1)/(2x^2 +4x -3) dx`
पुनः माना कि `2x+1 =A(d)/(dx) (2x^21 +4x -3) +B = A ( 4x+4)+B`
x के समान घाट को बरबा करने पर हमे मिलता है,
`4A=2" "rArr" "A =1/2" "` तथा `4A+B=1" "rArr" "B=-1`
अब, `I = int (1/2(4x+4)-1)/(2x^2+4-3)dx =1/2int (4x+4)/(2x^24x-3)dx-int (dx)/(2x^2""4x-3)+c=1/2 I_(1)-I_(2) +c" ...(1)"`
जहां ` I_(1) = int (4x+4)/(2x^2 +4x -3)dx ` तथा `I_(2) (dx)/(2x^2 +4x-3)`
`I_(1)` निकालना : `2x^2 +4x -3=t` रखे, तो ` (4x+4) dx = dt`
अब, `I_(1) int (dt)/t = log |t| =log |2x^2 + 4x-3|" ...(2)"`
पुनः `I_(2) int ( dx)/(""_((x+1)^2)-(sqrt(5/2))^2)=1/2""1/(2sqrt(5/2))log|(x+1-sqrt(5/2))/(x+1+sqrt(5/2))|`
`=1/(2sqrt(10))log|(x+1-sqrt(5/2))/(x+1+sqrt(5/2))|" ...(3)"`
(1), (2) तथा (3) से हममें मिलता है,
`I =1/2 log| 2x^2 +4x-3""log|(x+1-sqrt(5/2))/(x+1+sqrt(5/2))|+c`