फलन `int(3x+1)/(2x^(2)+x-1)dx` का x के सापेक्ष समाकलन कीजिए।

Explanation

माना `" "I=int(3x+1)/(2x^(2)+x-1)dx" …(1)"`
माना अंश `=A(d)/(dx)("हर")+B`
`rArr 3x+1=A(d)/(dx)(2x^(2)+x-1)+B`
`=A(4x+1)+B` ltBrgt `3x+1=4Ax+(A+B)`
दोनों पक्षों से समान घातीय पदों के गुणांकों की तुलना करने पर,
`3=4A rArr A=(3)/(4)`
तथा `A+B=1 rArr B=1-A=1-(3)/(4)=(1)/(4)`
अतः समीकरण (1 ) से ,
`I=int(A(d)/(dx)(" हर")+B)/(2x^(2)+x-1)dx`
`=(3)/(4)int((4x+1)dx)/(2x^(2)+x-1)+(1)/(4)int(dx)/(2x^(2)+x-1)`
`=(3)/(4)int((4x+1)dx)/(2x^(2)+x-1)dx+(1)/(8)int(dx)/(x^(2)+(1)/(2)x-(1)/(2))`
`=(3)/(4)log(2x^(2)+x-1)+(1)/(8)int(dx)/(x^(2)+(1)/(2)x_(1)/(16)-(1)/(2)-(1)/(16))`
`I=(3)/(4)log(2x^(2)+x-1)+(1)/(8)int(dx)/((x+(1)/(4))^(2)-((3)/(4))^(2))`
माना `x+(1)/(4)=t" "therefore" "dx=dt" तथा "(3)/(4)=a`
`therefore" "I=(3)/(4)log(2x^(2)+x-1)+(1)/(8)int(dt)/(t^(2)-a^(2))`
`=(3)/(4)log(2x^(2)+x-1)+(1)/(16a)log((t-a)/(t+1))`
`=(3)/(4)log(2x^(2)+x-1)+(1)/(16xx(3)/(4))log((x+(1)/(4)-(3)/(4))/(x+(1)/(4)+(3)/(4)))`
`=(3)/(4)log(2x^(2)+x-1)+(1)/(12)log((2x-1)/(2(x+1)))`