ज्ञात करें `int (x+1)/sqrt( 2x^2 +x-3) dx`

Explanation

माना कि `x+1 =A ["d.c. of " (2x^2 + x-3)+B]`
या, `x +1 =A(4x+1)+B = 4 Ax +A+B" ...(1)"`
x के समान घाट के गुणकों को बराबर करने पर हमे मोलता है ,
`4A=1" ":." "=1/4` तथा `A+B =1 " ":." "B=1 -A=1-1/4 =3/4`
अब, `I = int (x+1)/sqrt(2x^2 +x-3)dx = int (A(4x+1)+B)/sqrt(2x^2 +x-3)`
`A int (4x+1)/sqrt(2x^2 +x-3) dx +B int (dx)/sqrt(2x^2 +x-3)`
`=1/4 int (4x+1)/sqrt(2x^2+x-3)dx +3/4 int (dx)/sqrt(2x^2+x-3)" ...(2)"`
माना कि `I_1 =1/4 int (4x+1)/sqrt(2x^2 +x-3)`
`z=2x^2+-3` रखे, तो `dz = (4x+1) dx`
अब, `I_1 =3/4 int (dz)/sqrtz =1/4 int z^(-1/2) dz = 1/4""z^^(-1/2)/(1/2) = 1/2 sqrtz = 1/2 sqrt(2x^2 +x-3)`
माना कि `I_2 = 3/4 int (dx)/sqrt(2x^2 +x-3) =3/4 int (dx)/sqrt(2(x^2 + 1/2 x -3/2))`
`=3/(4sqrt2)int (dx)/sqrt(x^2 +2*x 1/4 +1/16 - 1/16 -3/2)`
`=3/sqrt(4sqrt2) int (dx)/sqrt((x+1/4)^2-25/16)`
माना कि `z =x +1/4,` तथा `a^2 =25/16` तो `1dz = dx`
अब, `I_(2)=3/(4sqrt2)int (dx)/sqrt(z^2 -a^2)log | z + sqrt ( z^2 -a^2)|`
`3/(4sqrt2) log| x+1/4+sqrt((x+1/4)^2-25/16)|+c`
`=3/(4sqrt2) log | x+1/4 + sqrt(x^2 +x/2-3/2)|`
`:." "` (2) से, ` I=1/2sqrt(2x^2 +x-3)+3/(4sqrt2)log| x+1/4 +sqrt(2x^2 +x-3)/sqrt 2|+c`