`int_(0)^(1)(x)/(x^(2)+1)dx`

Explanation

माना `I=int_(0)^(1)(x)/(x^(2)+1)dx`
माना`" "x^(2)+1=t" "rArr" "xdx = (dt)/(2)`
`" "x=0" "rArr" "t=1`
और `" "x=1" "rArr" "t=2`
`therefore" "I=int_(1)^(2)(1)/(t)(dt)/(2)=(1)/(2)int_(1)^(2)(1)/(t)dt=(1)/(2)[log|t|]_(1)^(2)`
`=(1)/(2)[log|2|-log|1|]=(1)/(2)log2`