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

Answer

.

Explanation

`int (2x+1)/sqrt(2x^(2)+x-3) dx = int (1/2 (4x+1)+1/2)/sqrt(2x^(2)+x-3) dx`
`=1/2 int (4x+1)/sqrt(2x^(2)+x-3) dx +1/2 int (1)/sqrt(2x^(2)+x-3) dx`
`=I_(1)+I_(2)` (मान लीजिए)
`I_(1)=1/2 int (4x+1)/sqrt(2x^(2)+x-3) dx`,
मान लीजिए `2x^(2)+x-3=t`
`(4x+1) dx=dt`
`=1/2 int (dt)/sqrt(t) =1/2 2t^(1//2) =sqrt(2x^(2)+x-3)`
`I_(2)=1/2 int (1)/sqrt(2x^(2)+x-3) dx`
`=1/(2sqrt(2)) int (1)/sqrt((x+1/4)^(2)-(5/4)^(2)) dx`,
मान लीजिए `x+1/4 =u`
`dx=du`
`=1/(2sqrt(2)) int (1)/sqrt(u^(2)-(5/4)^(2)) du`
`=1/(2sqrt(2)) log {u+sqrt(u^(2)-(5/4)^(2))}`
`=1/(2sqrt(2)) log {x+1/4+sqrt((x+1/4)^(2)-(5/4)^(2))}`
`=1/(2sqrt(2)) log {(4x+1+sqrt(16x^(2)+8x-24))/4}+c`
`=1/(2sqrt(2)) log {4x+1+sqrt(16x^(2)+8x-24)}-1/(2sqrt(2)) log 4+c`
`=1/(2sqrt(2)) log [4x+1+sqrt(16x^(2)+8x-24)]+c`, (जहाँ `c'=c-1/(2sqrt(2)) log 4`)
`:. int (2x+1)/sqrt(2x^(2)+x-3) dx=sqrt(2x^(2)+x-3)+1/(2sqrt(2)) log [4x+1+sqrt(16 x^(2)+8x-24)]+c`.