ज्ञात करें `int x/((x^2+1)(x+1))dx`

Explanation

माना की `int x/((x^2+1)(x+1))dx`
अब `x/((x^2+1)(x+1))=((x+1)^2-(x^2+1))/(2(x^2+1)(x+1))=1/2[(x+1)/(x^2+1)-1/(x+1)]`
`:." "I = 1/2[int (x+1)/(x^2+1) dx+int" "(dx)/(x+1)]+c`
`=1/2[int x/(x^2+1)dx + int (dx)/(x^2+1)-int (dx)/(x+1)]+c`
`=1/4log (x^2+1) +1/2 tan^(-1) x-1/2tan^(-1)x-1/2 log|x=1|+c`