ज्ञात करें `int (dx)/(3+2x-x^2), x lt 3`

Explanation

`3+3x-x^2 =-(x^2 -2x -3)=-(x^2 -3x+x-3)`
` = -[x (x-3)+1(x-3)]=- (x+1)(x-3)=(x+1)(3-x)`
अब `1/(3+2x -x^2) = 1/((x+1)(3-x))`
माना की `1/((x+1)(3-x))=A/(x+1)+B/(3-x)" ....(1)"`
`=(A(3-x)+B(x+1))/((x+1)(3-x))`
`:." "A(3-x)+B(x+1)=1`
`x=-1` रखने पर, `A* 4 =1 " ":." "A=1/4`
पुनः `x=3` रखने पर, `B*4=1" ":. B = 1/4`
अब, `I = int (dx)/(3+2x-x^2)=int (dx)/((x+1)(3-x)) = 1/4 int (1/(x+1)+1/(3-x))dx`
`=1/4 [log | x+1| - log | 3-x |] +c =1/4log|(x+1)/(3-x)|+c`