`int((5x-2))/((1+2x+3x^(2)))dx` का मान ज्ञात कीजिए ।

Explanation

`int((5x-2))/((1+2x+3x^(2)))dx`
अब माना कि `(5x-1)=A.(d)/(dx)(1+2x+3x^(2))+B`
तब `" "(5x-2)=A(6x+2)+B" …(1)"`
अब, समीकरण (1 ) के दोनों पक्षों के x की घातों की तुलना करने पर
`6A=5" तथा "2A+B=-2`
`rArr" "A=(5)/(6)" तथा "B=(-11)/(3)`
तब `" "int ({(5)/(6)(6x+2)-(11)/(3)})/((1+2x+3x^(2)))dx`
`=(5)/(6)int((6x+2))/((1+2x+3x^(2)))dx-(11)/(3)int(dx)/((3x^(2)+2x+1))`
`=(5)/(6)log(1+2x+3x^(2))-(11)/(3).(1)/(3)int(dx)/((x^(2)+(2)/(3)x+(1)/(3)))`
`=(5)/(6)log(1+2x+3x^(2))-(11)/(9)int(dx)/({(x+(1)/(3))^(2)+((1)/(3)-(1)/(9))})`
`=(5)/(6)log(1+2x+3x^(2))-(11)/(9)int(dx)/({(x+(1)/(3))^(2)+((sqrt2)/(3))^(2)})+c`
`=(5)/(6)log(1+2x+3x^(2))-(11)/(9).(1)/(((sqrt2)/(3)))tan^(-1){(x+(1)/(3))/((sqrt2)/(3))}+c`
`=(5)/(6)log(1+2x+3x^(2))-(11)/(3sqrt2)tan^(-1)((3x+1)/(sqrt2))+c`