ज्ञात करें `int(x^2+1)/((x+1)^2)dx`

Explanation

`z=x+1`, रखे तो `dz=dx` साथ ही `x=z-1`
अब `I=int(x^2+1)/((x+1)^2)dx=int((z-1)^2+1)/(z^2)dz`
`=int(z^2-2z+2)/z^2dz=(1-2/z+2/z^2)dz`
`=z-2log|z|-2/z+c=x+1-2log |x+1|-2/(x+1)+c`
`=x-2log|x+1|-2/(x+1)+k,` जहां `k=c+1`
Second method :
`int(x^2+1)/((x+1)^2)dx=int((x+1)^2-2x)/((x+1)^2)dx=int((x+1)^2-2(x+1)+2)/((x+1)^2)dx`
`=int[1-2/(x+1)+2/((x+1)^2)]dx=x-2log|x+1|-2/(x+1)+c`