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

Explanation

माना `" "I=int((x+1)dx)/(3-2x-2x^(2))" …(1)"`
माना `" "(x+1)=A(d)/(dx)(" हर ") +B`
`=A(d)/(dx)(3-2x-2x^(2))+B`
`=A(0-2-4x)+B`
या `" "(x+1)=-4Ax+(b-2A)`
दोनों पक्षों में समान घातीय पदों की तुलना करने पर,
`1=-4A rArr A=-(1)/(4)`
तथा `B-2A = 1 rArr B=2A+1=2(-(1)/(4))+1`
`therefore" "B=(1)/(2)`
अतः समीकरण (1 ) से,
`I=int(A(d)/(dx)("हर")+B)/(3-2x-2x^(2))dx`
`=-(1)/(4)int((-2-4x)dx)/(3-2x-2x^(2))+(1)/(2)int(dx)/((3-2x-2x^(2)))`
`=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dx)/((3)/(2)-(x^(2)+x))`
`=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dx)/(((3)/(2)+(1)/(4))-(x^(2)+x+(1)/(4)))`
`=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dx)/(((sqrt7)/(2))^(2)-(x+(1)/(2))^(2))`
माना `(sqrt7)/(2)=a` तथा `x+(1)/(2)=t" "therefore" "dx=dt`
`therefore" "I=-(1)/(4)log(3-2x-2x^(2))+(1)/(4)int(dt)/(a^(2)-t^(2))`
`=-(1)/(4)log(3-2x-2x^(2))+(1)/(4xx2a)log((a+t)/(a-t))`
`=-(1)/(4)log(3-2x-2x^(2))+(1)/(4xx2(sqrt7)/(2))log[(sqrt7+2x+1)/(sqrt7-2x-1)]`
`=-(1)/(4)log(3-2x-2x^(2))+(1)/(4sqrt7)log((sqrt7+2x+1)/(sqrt7-2x-1))`