ज्ञात करें `int (x)/((x^2 -a^2)(x^2-b^2))dx`

Explanation

माना की `z=x^2,` तो `2xdx = dz`
अब, `I =int (x)/((x^2 -a^2)(x^2 - b^2))dx =1/2 int(dz)/((z-a^2)(z-b^2))" ...(1)"`
माना की `1/((z-a^2)(z-b^2))=A/(z-a^2)+B/(z-b^2)" ...(2)"`
`=(A(z-b^2)+B(z-a^2))//((z-a^2)(z-b^2))`
`:." "A(z-b^2)+B(z-a^2)=1" ...(3)"`
`z=a^2` रखने पर, `A (a^2 -b^2)=1" ":." "A=1/(a^2 -b^2)`
`z =b^2` रखने पर, `B(b^2 - a^2) = 1 " ":." "B=-1/(a^2 -=b^2)`
`:." "` (1) से, `I = 1/2""1/((a^2 -b^2)) int (1/(z-a^2)-1/(z-b^2))dz`
` =1/(2(a^2 -b^2))[log |z-a^2|-log | z-b^2|] +c`
`=1/(2(a^2-b^2))log|(z-a^2)/(z-b^2)|+c=1/(2(a^2-b^2))log|(x^2-a^2)/(x^2-b^2)|+c" "[:." "z=x^2]`