`(5x-2)/(1+2x+3x^(2))`

Explanation

`int(5x-2)/(1+2x+3x^(2))dx=I` (माना)
माना `5x-2=A(d)/(dx)(1+2x+3x^(2))+B`
`rArr 5x-2=A(2+6x)+B`
x गुणांक बराबर रखने पर
`5=6A" "rArr" "A=(5)/(6)`
अचर पद बराबर रखने पर,
`-2=2A+B`
`rArr" "-2=2xx(5)/(6)+B`
`rArr" "B=-(11)/(3)`
`therefore" "I=int((5)/(6)(2+6x)-(11)/(3))(1+2x+3x^(2))dx`
`=(5)/(6)int(2+6x)/(1+2x+3x^(2))dx-(11)/(3)int(1)/(1+2x+3x^(2))dx`
`=(5)/(6)int(2+6x)/(1+2x+3x^(2))dx-(11)/(9)int(1)/(x^(2)+(2)/(3)x+(1)/(2))dx`
माना समाकलन के लिये
माना `1+2x+3x^(2)=t`
`rArr" "(2x+6x)dx=dt`
`=(5)/(6)int(1)/(t)dt-(11)/(9)int(1)/((x^(2)+(2x)/(3)+(1)/(9))+((1)/(3)-(1)/(9)))dx`
`=(5)/(6)int(1)/(t)dt-(11)/(9)int(1)/((x+(1)/(3))^(2)+((sqrt2)/(3))^(2))dx`
`=(5)/(6)log|t|-(11)/(9).(1)/(sqrt2//2)tan^(-1)((x+(1)/(2))/(sqrt2//3))+c`
`=(5)/(6)log|1+2x+3x^(2)|-(11)/(3sqrt2)tan^(-1)((3x+1)/(sqrt2))+c`